If $B$ is the Borel sigma algebra on a metric space $X$ and $\{F_a\}$ is a family of closed subsets of $X$, then $\bigcap_a F_a \in B$.

Question

I conjectured the following statement and attempted a proof. Can you please verify whether my proof is correct?

If $B$ is the Borel sigma algebra on a metric space $X$ and $\{F_a\}$ is a family of closed subsets of $X$, then $\bigcap_a F_a \in B$.

Proof. Let $\mathcal{T}$ denote the collection of open subsets of $X$, i.e., the topology on $X$. Let $\{F_a : a \in A\}$ be a collection of closed subsets of $X$. Then $F_a^c \in \mathcal{T}$ for all $a \in A$. But $\mathcal{T}$ is a topology, so $\bigcup_{a\in A}F_a^c \in \mathcal{T}$. The Borel sigma algebra is the sigma algebra generated by the open sets in $\mathcal{T}$, so $\bigcup_{a\in A}F_a^c \in B$ and therefore $\bigcap_{a\in A} F_a = \left(\bigcup_{a\in A}F_a^c\right)^c \in \mathcal{T}$.

Related question

Does the statement hold if we replace $\bigcap_{a\in A} F_a$ by $\bigcup_{a\in A} F_a$?