(For convenience of exposition, all segments mentioned below do not include their two endpoints.)
For any line in the plane, we say it is "red" if and only if it contains two points of $M$. Since no three points of $M$ are collinear, each red line uniquely determines two points of $M$. Moreover, the number of red lines is clearly $C_2^n < n^2/2$. Furthermore, for each segment, let its "intersection number" be the number of red lines that intersect it. For a set of segments, its "intersection number" is then defined as the sum of the intersection numbers of the individual segments. We will prove that
1. At the start of the game, the intersection number of all the segments is less than $n^3/2$.
2. Each time an operation is performed, the intersection number of all the segments decreases by at least 2.
Since the intersection number of all the segments can never be negative (otherwise the operation could not be performed), the above two points together prove the original claim.

Pf of 1. Since each segment's intersection number is less than $C_2^n < n^2/2$, and there are $n$ segments in total, the claim follows.

Pf of 2. Suppose we choose $AB$ and $CD$, and after the operation change them to $AC$ and $BD$. Let $X_{AB}$ be the set of all red lines that intersect $AB$, and define $X_{CD}$, $X_{AC}$, and $X_{BD}$ similarly. To prove the second point, it suffices to prove that
$$|X_{AC}| + |X_{BD}| + 2 \le |X_{AB}| + |X_{CD}|, \quad (1)$$
where $|X|$ denotes the number of lines in $X$.

First, if a line intersects $AC$, then it must intersect $AB$ or $CD$. Furthermore, the line $AB$ is a red line for the segment $CD$, and the line $CD$ is a red line for the segment $AB$, but neither of them is a red line for $AC$ or $BD$ (since they meet at endpoints). Hence, we have
$$|X_{AC} \cup X_{BD}| + 2 \leq |X_{AB} \cup X_{CD}|. \qquad (2)$$
On the other hand, if a line is simultaneously a red line for both $AC$ and $BD$, then it must inevitably intersect both $AB$ and $CD$. Hence
$$|X_{AC} \cap X_{BD}| \leq |X_{AB} \cap X_{CD}|. \qquad (3)$$
Combining (2) and (3), we obtain (1), completing the proof.