Proof that an intersection of closed sets of a metric space is closed

Proof

1. Let $(X, d)$ be a metric space, let $I$ be a nonempty set, and let $(F_i)_{i \in I}$ be a family of closed subsets of $X$.
2. Put $F = \bigcap_{i \in I} F_i$, the intersection of the family.
3. We claim that $X \setminus F = \bigcup_{i \in I} (X \setminus F_i)$, a union of complements.
4. Let $x \in X$. Then $x \notin F$ if and only if $x \notin F_j$ for some $j \in I$.
5. This holds if and only if $x \in X \setminus F_j$ for some $j \in I$, that is, $x \in \bigcup_{i \in I} (X \setminus F_i)$. This proves the claim.
6. Each $F_i$ is closed, so each $X \setminus F_i$ is open, by the definition of a closed set.
7. By union of open sets is open, the set $\bigcup_{i \in I} (X \setminus F_i)$ is open. By step 3, $X \setminus F$ is open.
8. By the definition of a closed set, $F$ is closed in $X$.