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

Let $T$ be a maximal set such that $T \subseteq S$ for infinitely many $S \in \mathscr{F}$, which form $\mathscr{F}' \subseteq \mathscr{F}$ (such $T$ exists, since the empty set works).

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

But $U$ s-meets $\mathscr{F}'$, so by pigeonhole, there must exist $u \in U \setminus T$ belonging to infinitely many $S \in \mathscr{F}'$, contradicting the maximality of $T$.

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