User:DanZimm/Sandbox

If $x \in \bigcap_{i \mathop \in I} \bigcap \mathbb S_i$ then $\forall i \in I \; \forall A \in \mathbb S_i \; x \in A$; which means that for every $A$ in at least one of $\mathbb S_i$ we have $x \in A$. This is equivalent to saying $\bigcap_{i \mathop \in I} \bigcap \mathbb S_i \subset \bigcap \bigcup_{i \in I} \mathbb S_i$. For the opposite inclusion note that if $x$ is in every $A$ that is in at least one of $\mathbb S_i$ then for every $i$, for every $A \in \mathbb S_i \; x \in A$.

If $x$ is in $A$ for every $A$ in any $\mathbb S_i$ if and only if $x$ is in $A$ for every $A$ that's in some $\mathbb S_i$. This is the crux.