SAUDI ARABIAN MATHEMATICAL COMPETITIONS

We call a set $S$ s-meets $\mathcal{F}$ if it shares at least $s$ elements with each set in $\mathcal{F}$. Suppose no such set of size (at most) $r-1$ exists. (Each $S\in \mathcal{F}$ s-meets $\mathcal{F}$ by the problem hypothesis.)

Let $T$ be a maximal set such that $T\subseteq S$ for infinitely many $S\in \mathcal{F}$, which form ${\mathcal{F}}^{\prime }\subseteq \mathcal{F}$ (such $T$ exists, since the empty set works).

Clearly $|T|<r$, so by assumption, $T$ does not s-meet $\mathcal{F}$, and there exists $U\in \mathcal{F}$ with $|U\cap T|\le s-1$.

But $U$ s-meets ${\mathcal{F}}^{\prime }$, so by pigeonhole, there must exist $u\in U\setminus T$ belonging to infinitely many $S\in {\mathcal{F}}^{\prime }$, contradicting the maximality of $T$.

Remark. Let $X$ be an infinite set, and $a_1,a_2,\dots ,a_{2r-2-s}$ elements not in $X$. Then the set $$\mathcal{F}=\left\{B\cup \{x\}:B\subseteq \{a_1,a_2,\dots ,a_{2r-2-s}\},|B|=r-1,x\in X\right\}$$ shows we cannot replace $r-1$ with any smaller number.