Let $E$ be the set of the edges made by those points. We will say that $x<<y$ for some $x,y\in E$ iff there is such an increasing path between $x$ an $y$. It is clear that $(E,<<)$ is a poset. Obviously any two edges that share a common vertex are comparable, therefore the maximum number of elements in an antichain is $n/2$, so any partition of $E$ in antichains has at least $2q/n$ antichains. Now we just have to use Mirsky Theorem and we are done.

Unfortunately the above solution is not correct as the existence of a chain $a < b < c < \cdots $ does not guarantee a polygonal line through the corresponding segments. Consider for example the case in which three line segments meet at a single point. Then all edges are comparable, but we can only form a polygonal line using two of these segments. I cannot see how to rectify the above by using the same idea but with directed segments (we may have antichains consisting of more than $n$ elements). Here is a different approach that works: Let $\ell_1 < \ell_2 < \cdots < \ell_q$ be the lengths of the segments. We direct all segments once in each direction and consider these $2q$ segments. For each $1 \leqslant r \leqslant q$ we will construct a set $L_r$ of at most $n$ directed polygonal lines of increasing length with the following properties: (a) We will use all segments of length $\ell_i$ for $1 \leqslant i \leqslant r$ (and only those) once in each direction. (b) Each of these directed polygonals lines will end at a different point. The case $r=1$ is immediate. Suppose the case $r = k < q$ is true. Let $L_k$ be the set of directed polygonal lines guaranteed by the induction hypothesis, and let $AB$ be the segment of length $\ell_{k+1}$. We define $L_{k+1}$ as follows: - If a directed polygonal line belongs to $L_k$ and ends at a point different from $A$ and $B$ then we add it to $L_{k+1}$. - If there is a directed polygonal line that belongs to $L_k$ and ends in $A$ then we extend it by the segment $AB$ and add it to $L_{k+1}$. (There is at most one such by (b)) Otherwise, we add the directed polygonal line $AB$ to $L_{k+1}$. - Analogously, if there is a directed polygonal line that belongs to $L_k$ and ends in $B$ then we extend it by the segment $BA$ and add it to $L_{k+1}$. (There is at most one such by (b)) Otherwise, we add the directed polygonal line $BA$ to $L_{k+1}$. - $L_{k+1}$ contains no other directed polygonal lines than the ones described above. We now observe that (a) and (b) are still true for $L_{k+1}$ and so the case $r=k+1$ also holds. So our claim is true by induction. For $r=q$ we have at most $n$ directed polygonal lines consisting of a total of $2q$ segments so one of them uses at least $2q/n$ segments. Ignoring the directions we now have our polygonal line that we want.

^ @Demetres: Can you please fix up your solution, it is not written clearly. For example I am left confused by the following (among other) things: (1) Fix up condition (a), what does "We will all segments of length..." really mean? What happens if the edges of length $\ell_1$ and $\ell_2$ do not share a common vertex at all, then how can such a path (of length exactly $r$ I presume) exist? (2) How does your induction step even use the crucial fact that $ r = k <\lceil 2q/n\rceil $? (3) What does "We perform exactly the same operation..." mean? What operation?

Apologies, I have now edited my post. Hopefully it is clearer now. Regarding your comments: (1) The word "use" was missing from the statement. My paths (directed polygonal lines) do not have length r. There are many paths not just one with the sum of their lengths being equal to 2r. If the edges of lengths $\ell_1$ and $\ell_2$ (say $AB$ and $CD$) do not share a common vertex then at the second step of the construction I will have four directed paths. Namely, $AB,BA,CD$ and $DC$. (2) Oops, that should have been $r=k < q$. (3) I thought that was clear but I have rewritten the whole paragraph into what I believe it is more transparent now.

Great, I understand now. Nice solution.

Consider the direct graph where we each edge has two directions ($2q$ edges). A directed edge $e$ is called terminal if there isn't an edge $f$ with larger value such that the end vertex of $e$ is the same as the start vertex of $f$. At step $i$, we remove all terminal edges in the remaining graph. If we can show that each step removes at most $n$ direct edges, then $2q/n$ steps are required to remove all edges, implying there exists a polygonal line of increasing length of length at least $2q/n$ Let $I_i(a), O_i(a)$ be the in-degree and out-degree of $a$ before step $i$. Assume we are at the time before step $k$. Put $m = O_k(a)$, and let $e$ be the outgoing edge of $a$ with the largest value $v$. Then $v \geq m$. After step $k$, only incoming edges of $a$ with value $ \geq m$ will have possibly been removed. Thus $I_{k+1}(a) \geq m-1 = O_k(a)$. Sum it over all $a$ \[\#\text{edges before step k+1} = \sum_a I_{k+1}(a) \geq \sum_a O_k(a) -n = \#\text{edges before step k}\] and we have shown the number of edges after each step decreases by at most $n$.