I was asked (by myself) to give a proof of the following seemingly simple geometric statement, but after thinking a little I now suspect it could be less elementary than I thought (or am I being silly?). Does anybody know it, and can give an answer or a reference to it? Of course, I'm quite sure it should fit within a larger theory in combinatorics or in probability, but an elementary answer would be appreciated.
Let $S$ be a (say open) subset of a square $[0,1]^2$ with Lebesgue measure $|S|>1/2$. Then, there exists a rectangle with a vertex on the diagonal, and the other three vertices in $S$ (in other words, there are three points of $S$ of the form $(x,y)$, $(x,z)$ and $(y,z)\ $).
The constant $1/2$ cannot be lowered, as the example of the subset $S^* :=(0,1/2)\times(1/2,1)\cup(1/2,1)\times(0,1/2)$ shows (for any three points $x,y,z$ in $[0,1]$, at least 2 of them are both either smaller or larger than $1/2$, so the corresponding pair is not in $S^*$.