The idea is to translate the claim into terms of the complements of the sets in the given family. Denote $$\mathcal{F}' := \{S\setminus F : F\in \mathcal{F}\}$$Using De Morgan laws, the family $\mathcal{F}'$ satisfies the following conditions 1) $\mathcal{F}'\setminus \{\emptyset\}\neq \emptyset $ 2) If $F_1,F_2\in\mathcal{F}'$ then $F_1\cap F_2\in \mathcal{F}'$ and $F_1\cup F_2\in \mathcal{F}'$ We want to prove there exists $a\in S$ which belongs to at least half of the elements of $\mathcal{F}'$. Let $A\in\mathcal{F}'$ be a minimal element (under inclusion) of the family $\mathcal{F}'\setminus \{\emptyset\}.$ By 1) it's guaranteed $A\neq \emptyset.$ Any other set in $\mathcal{F}'$ is either disjoint from $A$ or contains $A.$ Let $a\in A$. We prove $a$ belongs to at least half of the elements of $\mathcal{F}'$. Consider $\mathcal{F}'_1:= \{F\in\mathcal{F}' : F\cap A=\emptyset\}.$ All sets of $\mathcal{F}' $ that does not contain $a$ are exactly the sets in $\mathcal{F}'_1$. Consider the subfamily $\mathcal{F}'_2:=\{A\cup F :F\in \mathcal{F}'_1\}.$ They all contain $a$ and $|\mathcal{F}'_2|=|\mathcal{F}'_1|$. The result follows.