Let $A$ be a $\sigma$-algebra on $X$ and let $\{B_n\} \subset A$. Define $F = \{x \in X$: there exist infinitely many $n$ such that $x \in B_n$$\}$.
If the number of $B_n$ is countable, I can prove that $F$ is in $A$. But what if there are uncountably many $B_n$? Does the result change? I cannot produce a counterexample (though that is largely because I have no intuition regarding $\sigma$-algebras).
Here is a counterexample. Let $X$ be any uncountable set and let $A$ be the $\sigma$-algebra of subsets of $X$ that are either countable or cocountable. Let $F\subset X$ be any subset that is neither countable nor cocountable. Now let $\{B_i\}$ be the set of all countable subsets of $F$. Then each $B_i$ is in $A$ and $F$ is exactly the set of elements of $X$ that are in infinitely many of the $B_i$, but $F$ is not in $A$.