Each of 20 students has made a choice of anywhere from 0 to 6 courses from a total of 6 courses offered. Prove or disprove: there are 5 students and 2 courses such that all 5 have chosen both courses or all 5 have chosen neither course.
The statement is false and you can look up the proof. However, I have an argument that "proves" it is true, please tell me what I'm doing wrong.
Assume that for any two courses, fewer than 5 students have chosen both courses. Let $C_1$ be the most chosen course, by $k$ students namely $S_1,\dots, S_k$. Any course other than $C_1$ could have been chosen by at most 4 of $S_1,\dots, S_k$ and at most $20-k$ of $S_{k+1},\dots, S_{20}$. Thus, $k\geq (20-k)+4$ or $k\geq12$. Now, we can easily see that there are 5 students of $S_1,\dots, S_k$ who didn't choose two of $C_2,\dots, C_5$.
Community wiki answer so the question can be marked as answered:
As discussed in the comments, you can’t conclude that $k\geq (20-k)+4$ merely from the fact that at most $(20-k)+4$ students have chosen any other course. To conclude this, there would actually have to be a course that achieves that (very loose) bound.