TITLE: $551$ rectangles are selected from a $10\times 10$ square. Prove that one of these rectangles is inside another one (they can have common side).
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: Proof for $386$.
For $R,S$ rectangles, define $R\sim S$ if and only if all these three requirements holds:

$R\subseteq S$ or $S\subseteq R$;
$R,S$ have the same left-down corner, say at position $(x,y)$;
if $y\leq x$ then $R,S$ have the same base, otherwise the same height.

Then $\sim$ is an equivalence relation, its equivalence classes are chain of rectangles.
The number of equivalence classes is given by
$$\sum_{n=1}^{10}\sum_{m=1}^{10}\min\{n,m\}=385$$
because for each pair $(n,m)$ there are exactly $\min\{n,m\}$ equivalence classes whose rectangle have left corner at position $(10-n,10-m)$.
Since an antichain shares with each equivalence class at most a rectangle, it follows that each antichain contains at most $385$ rectangles.