Suppose it's not true. Call polygon as $A_1A_2.....A_{2n-1}$ and points are in anticlockwise order. Now call each of pair of points $(A_kA_{2n+1-k}),k>2$ good pair and inverse one of another, means $A_k$ is inverse point of $A_{2n+1-k}$ or opposite also . Now Let $A_1$ is chosen, as there are $n-1$ distinct good pair so we can take exactly one point from each of the good pairs. Now WLOG let $A_1$ is chosen so clearly we can't take $A_{n+1}$ implies we've to take $A_{n}$. As $A_1,A_2$ are consecutive vertices so one of $A_1A_n$ or $A_2A_n$ has odd points at RHS. Consider that pair , so for that we'll get $m>n$ such that we can't take $A_m$, so we've take it's inverse point which is at LHS of $A_n$ call $A_x$ with $x<n$ similarly again we'll get $A_y$ with $y<x<n$ so going on we must have to take $A_3$ (because that sequence won't go to $A_2$ or $A_1$ before going to $A_3$) but then $A_1A_2A_3$ is Isosceles. Hence a contradiction.

For choice of a set $S$ of $n$ vertices, call one of the other $n - 1$ vertices bad if it would form an isosceles triangle with some pair of elements in $S$. Since the polygon has an odd number of sides, for each pair of vertices there is a unique vertex which would form an isosceles triangle with the pair we chose; we will say the pair forbids this vertex. For the sake of contradiction, suppose no three vertices in $S$ form an isosceles triangle. Then each of the $n$ vertices cannot be forbidden by any pair of vertices in $S$. There are \[\binom{n}{2} = \frac{n(n - 1)}{2}\] pairs of elements in $S$ so the total number of times some vertex is forbidden is just that. But since no vertex in $S$ is forbidden, the remaining $n - 1$ vertices must together be forbidden a total of $\frac{n(n - 1)}{2}$ times, meaning some vertex must be forbidden at least $\frac{n}{2}$ times. It is quite easy to see that given a forbidden vertex and one of the two vertices that forbids it, the other is uniquely determined. Then a single vertex is forbidden at most $\frac{n}{2}$ times by a set of $n$ vertices. Thus equality must hold throughout meaning for each of the $n - 1$ vertices $v$ not in $S$ (a set we will denote by $S'$), $S$ is symmetric with respect to $v$. No two vertices in $S'$ can be consecutive, since some vertex in $S'$ would have an element of $S$ and an element of $S'$ adjacent to it, contradiction. Then vertices must alternate $S$, $S'$, $S$, $\dots$ with exactly one pair of vertices $(v_1, v_2)$ in $S$ adjacent. But then we can choose a vertex $v$ in $S'$ so that both these vertices are on the same side of the dividing line through $v$, contradiction.

Is this correct ? Assume that it is not true Assume that $A_1,A_2,...,A_{2n-1}$ are the vertices of our polygon , $A_{2n+k}=A_{k+1}$ , $k$ is positive Define : $l_i$ is the line throw $A_1$ and perpendicular to $A_{n-1+i}A_{n+i}$ Clearly $A_{i-k}$ is the reflection of $A_{i+k}$ across $l_i$ and clearly there is no three consecutive points. Assume we chose $A_i$ , we have$ 2(n-1) $ points without $A_i$ If we chose $A_{i+k}$ then we can't choose $A_{i-k}$ because of symmetry , so atmost we can choose $n-1$ points but in fact we'll choose $n-1$ points . Assume we choose $A_1$ , then we must choose one of $A_2$ or $A_{2n-1}$ , WLOG assume we chose $A_{2n-1}$ , we can't choose $A_{2n-2}$ , because can't choose three consecutive points , then we must choose $A_{3}$ draw $l_3$ so we must choose $A_4$ since we can't choose $A_2$ , draw $l_{2n-1}$, we must choose $A_{2n-3}$ since we can't choose $A_2$ , but this implies that we chose $A_1,A_4,A_{2n-3}$ . Contradiction !!

If all of chosen $n$ points are distanced exactly $2$ from each other it is trivial.Suposse they are not.That means that there are $2$ points that are next to each other.Now suposse contrary,that we can choose $n$ points in such maner and let those points be: $A_1,A_2, \cdots , A_n$ and such that $A_1$ is next to $A_2$.Now we see that we can divide any $2n-2$ points in pairs such that points from the same pair are equaly distanced from 2n-1-th point.Note that in every pair one point will be chosen and the other won't.That means that points $A_2,A_3, \cdots , A_n$ all belong to different pairs with respect to $A_1$.That also hold for $A_2$.Because points $A_3,A_4, \cdots , A_n$ are the same in both cases they form different pairs wrt different points.Now we just draw it and see what happens because of conditions(condition that one point in a pair determines the other and all points belong to $2$ pairs).We then trivialy get a contradiction so it is proven.

Ignore wrong solution

Not that, as you say, it is "difficult to understand" (its intention is in fact quite easy to grasp), but it's incorrect. You try to create $n-2$ "pigeonholes" (pairs of vertices), where to place $n-1$ "pigeons" (the remaining vertices), but of course you only account for $2(n-2)+1 = 2n-3$ vertices. In fact there are $n-1$ "pigeonholes"; the last is not your $\{A_{k+n-2}, A_{k+n-1}\}$, which is not valid, but $\{A_{k+n-1}, A_{k+n}\}$.

SmartClown wrote: If all of chosen $n$ points are distanced exactly $2$ from each other it is trivial. Suposse they are not. That means that there are $2$ points that are next to each other. The preamble is unnecessary; any choice of $n$ points among $2n-1$ on a circle always contains two neighbouring points. But the idea is good, and allows a quick finish. Let us index the $2n-1$ points modulo $2n-1$. Say $A$ is the set of indices of the chosen $n$ points. For an arbitrary fixed $i\in A$ let us consider the $n-1$ doubletons $\{i-k, i+k\}$, for $1\leq k \leq n-1$. If any other two indices from $A$, distinct from $i$, belong to a same doubleton, an isosceles triangle is created. So assume this never happens, for any $i\in A$, thus the $n-1$ indices from $A\setminus \{i\}$ each belongs to a different doubleton. WLOG assume $A$ contains $0$ and $1$, as argued above. For any $i\in A$ and $0\leq j\neq i \leq 2n-2$ denote by $j \in A \stackrel{(i)}{\longrightarrow} 2i-j \not \in A$ or $j \not \in A \stackrel{(i)}{\longrightarrow} 2i-j \in A$ the fact that exactly one of $j,2i-j$ belongs to $A$. Then $1 \in A \stackrel{(0)}{\longrightarrow} -1 \not \in A \stackrel{(1)}{\longrightarrow} 3 \in A \stackrel{(0)}{\longrightarrow} -3 \not \in A \stackrel{(1)}{\longrightarrow} 5 \in A$, and so $\{1,3,5\} \subseteq A$, contradiction with $1 \stackrel{(3)}{\longrightarrow} 5$. Therefore the assumption was false, and an isosceles triangle is always created.

As a matter of fact, working with $0,1 \in A$ is not instrumental. Let us do the same for any $i,j \in A$ with $i\neq j$ (under the notations of above; the idea is reminiscent of one in the proof that a finite set cannot have two centres of symmetry, alternately taking symmetrics with respect to them). Then \[j \in A \stackrel{(i)}{\longrightarrow} 2i-j \not \in A \stackrel{(j)}{\longrightarrow} 3j-2i \in A \stackrel{(i)}{\longrightarrow} 4i-3j \not \in A \stackrel{(j)}{\longrightarrow} 5j-4i \in A,\] and so $\{j,3j-2i,5j-4i\} \subseteq A$, contradiction with $j \stackrel{(3j-2i)}{\longrightarrow} 5j-4i$. For the record, the indices $j$, $3j-2i = j + 2(j-i)$, $5j-4i = j + 4(j-i)$ are distinct, since $2n-1 \nmid 2^k(j-i)$ for any $k\geq 0$, because $2n-1 \nmid j-i$. Therefore the assumption was false, and an isosceles triangle is always created.

Label the points $-n+1, -n+2, ..., -1, 0, 1, 2, ..., n-1$, let $V$ be the set of the $n$ points. WLOG, suppose $0 \in V$. Then, we make the following observations: 1. At most one of $k, -k \in V$. Suppose otherwise, then the points $-k, 0, k$ form an isosceles triangle. 2. Exactly one of $k, -k \in V$. There are $n-1$ pairs of $k, -k$, but a total of $n-1$ remaining terms in $V$. Using the first observation, we can conclude that exactly one of $k, -k \in V$. It is safe to assume $1 \in V$. If $1 \in S$, $-1, 2 \not\in V$. $2 \not \in V \implies -2 \in V$. $0, -2 \in V \implies -4 \not\in V$. $-2, 1 \in V \implies 4 \not\in V$. But this contradicts observation 2, hence a contradiction for $n \geq 5$. For $n = 3$ or $4$, one can easily check that an isosceles triangle will always be formed.