$551$ rectangles are selected from a $10\times 10$ square. Prove that one of these rectangles is inside another one (they can have common side).

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I'm actually looking for a solution of extension of this problem:

We have a set $$M = \{(x,y)\in \mathbb{Z}^2; 0\leq x,y\leq 10\}$$

Let $P$ be a subset of rectangles with vertices in $M$ and with sides parallel to coordinate axis. What is the smallest cardinality of $P$ that we can find for sure in $P$ two rectangles $R$ and $R'$ such that $R\subseteq R'$?


Proof for $551$: Every rectangle is uniquely determined with two parallels with x-axis and two parallel with y-axis. Since we have $11$ parallel lines with x-axis, we can chose those on ${11\choose 2}=55 $ ways. So we have two parallels to x-axis with at least $11$ rectangles with sides on these two parallels.

Now let us observe only these $11$ rectangles. Each is uniquely determined with their left and right side. Since we have $11$ rectangles and only $10$ possibilities we have two of them with left side on the same parallel to y-axis. So one of those includes the other one and we are done.