Problem $2$ from USAMO, $1989$: The $20$ members of a local tennis club have scheduled exactly $14$ two-person games among themselves, with each member playing in at least one game. Prove that within this schedule there must be a set of six games with $12$ distinct players.
My attempt at a solution: Since there are $20$ players, each of whom has played at least one game, the least number of matches which must be organised to accommodate all players would be $10$. Now, $4$ more matches are left, which can be played by at most $8$ of these $20$ players. Thus, at least $12$ of the players get to play no more than $1$ match, and hence there must exist a set of six games with $12$ distinct players. This, I believe, will be the guiding principle of a rigorous proof.
I have the following doubts clouding my mind:
- Is my reasoning sound, i.e., free of any pitfalls?
- If correct, how to frame the above reasoning in a mathematically rigorous proof?
Edit: As pointed out by Ben in the comments, the statement in italics is unjustified. I overlooked a lot of possibilities while framing this proof, it seems. Thus, I would like to get some hints to proceed with more reasonable proof.