Let $AB$ be a chord, $f(AB)$ is defined as the chord (if there are a lot just pick an arbitrary one) whose midpoint is passed by $AB$. Suppose that a chord $A_1B_1$ passes through the midpoint $M$ of the chord $A_2B_2$. $A_1B_1=A_1M+B_1M\geq2\sqrt{A_1M\times B_1M}=2\sqrt{A_2M\times B_2M}=A_2M+B_2M=A_2B_2$, and the equation holds $\iff A_1M=B_1M\iff A_1A_2B_1B_2$ is a parallelogram $\overset{\angle A_1A_2B_1+\angle B_1B_2A_1=180^o}\iff A_1A_2B_1B_2$ is a rectangle $\iff A_1B_1, A_2B_2$ are diameters. From the above, if there is a chord $AB$ that is not a diameter, then $AB>f(AB)$, and $f(AB)$ is not a diameter either. $\Rightarrow AB>f(AB)>f^2(AB)>\cdots$, and $C\supset\{AB, f(AB), f^2(AB), \dots\}$, which is not a finite set. Therefore, there does not exist a chord that is not a diameter, and this means that any two of these chords intersect inside the circle.

Same as above probably. Let $AB$ be a chord with midpoint $M$ and $CD$ another chord passing through $M.$ Then $|CD| \geq |AB|$ since $|CD| = |CM| + |DM| \geq 2 \sqrt{ |CM| \cdot |DM| } = 2 \sqrt{ |AM|^2 } = 2 |AM| = |AB|,$ by power of a point. Equality holds iff $|CM| = |DM|,$ but this can only occur if $AB$ and $CD$ are both diameters. Consider a directed graph where each chord is represented by a vertex, and $c_i \to c_j$ if chord $c_1$ passes through the midpoint of $c_2.$ Thus the outdegree of each vertex is at least one. Then starting from any vertex take a walk, $c_1 \to c_2 \to \cdots$ Since there are at most $n$ vertices after at most $n+1$ steps some vertex will be repeated, say $c_k$ is the first repeated vertex. Then there is a cycle from $c_k, c_{k+1}, \ldots, c_l, c_k,$ but this means $|c_k| \geq |c_{k+1}| \geq \cdots \geq |c_l| \geq |c_k|,$ so equality holds along this cycle and all of these chords are diameters. Finally, since there is a path from $c_1$ to $c_k$ we have $|c_1| \geq |c_k|$ and hence $c_1$ is also a diameter. This reasoning holds for any starting chord so thus all chords $c_i$ are diameters, and thus all chords intersect at the center of the circle.