Given a positive integer $n \geq 5$ and a convex polygon $P$ with vertices $A_1, A_2, \ldots, A_n$, we need to prove that it is possible to choose a point inside every quadrilateral $A_iA_jA_kA_l$ (where $1 \leq i < j < k < l \leq n$) such that the chosen points are distinct and any segment connecting these points intersects some diagonal of $P$.

Define an intersection point to be the intersection of two diagonals within $P$. If no three diagonals of $P$ are concurrent, then each intersection point $X$ is uniquely defined by the intersection of two diagonals $A_iA_k$ and $A_jA_l$, so $X$ is in the interior of quadrilateral $A_iA_jA_kA_l$.

The diagonals of $P$ divide it into several regions. We wish to show that each intersection point $X = A_iA_k \cap A_jA_l$ may be assigned a unique region $R_X$ touching it. If this assignment is possible, then $R_X$ is contained in the quadrilateral $A_iA_jA_kA_l$, and we choose a point within $R_X$. The $\binom{n}{4}$ chosen points will all lie in different regions, so any segment connecting two of the chosen points must intersect some diagonal of $P$.

Lemma. If some (at least one) intersection points are colored blue, there is a region containing exactly one blue point on its perimeter.

Proof of Lemma. Note that each intersection point touches exactly four regions. Suppose each of the four regions $R_1, R_2, R_3, R_4$ touching blue point $X$ have another blue vertex $X_1, X_2, X_3, X_4$ (all distinct from $X$, but there may be repeated points among them). If we extend the diagonals through $X$ to infinite lines, the entire plane is divided into four sectors, each containing $X_1, X_2, X_3, X_4$ respectively (they may lie on the boundaries of the respective sectors). Therefore, $X$ lies in the convex hull of $X_1, X_2, X_3, X_4$.

Thus, pick a blue point $X$ on the convex hull of the set of all blue points. It must touch a region that has no blue point other than $X$, which proves the Lemma. $\square$

Initially, color all $\binom{n}{4}$ intersection points blue. Then, repeatedly apply the Lemma to find region $R_X$ with sole blue vertex $X$; assign $X$ to $R_X$, and remove the color from $X$. Eventually, each intersection point is assigned a region touching it. If two intersection points $X$ and $Y$ were assigned the same region $R_X = R_Y$, where without loss of generality $X$ was assigned the region first, then $Y$ would have been a second blue vertex of $R_X$, contradiction. Therefore, the assigned regions are unique, and the question statement follows.

The answer is: \boxed{\text{Proven}}.