Let $X$ be a metric space and $A \subseteq X$ open. For $n \in \mathbf{N}$, define the set $$ A_n := A \cap (B(0, n) - \overline{B(0, n - 1)}) $$ with $B(0, 0) = \emptyset$. Suppose for each $n \in \mathbf{N}$, there exists $F_n \subseteq A_n$ closed. I would like to show $\bigcup_{n = 1} ^\infty F_n$ is closed. I am a bit surprised, in fact, that this is true. Any suggestions?
My Attempt: Let $\{ x_n \}_{n = 1} ^\infty$ in $\bigcup_{n = 1} ^\infty F_n$ be given such that $x_n \to x$. The goal is to show $x \in \bigcup_{n = 1} ^\infty F_n$. Now for $\epsilon > 0$, there exists $N_\epsilon \in \mathbf{N}$ large such that $d(x, x_{N_\epsilon}) < \epsilon$. This gives $x \in B(0, \tilde{N}(\epsilon))$ for some $\tilde{N}(\epsilon)$ large. Then $x \in A \cap \bigcup_{n = 1} ^{\tilde{N}_\epsilon} B(0, n) - \overline{B(0, n - 1)} = A_n$. I am a bit stuck from here on.
A couple things I've noticed:
The definition of $A_n$ is mutually disjoint and thus $F_n$ is also mutually disjoint.
Each of $A_n$ is an open set.
To converge, $\{x_n\}$ must eventually end up in at most two of the $A_n$, and thus in at most two of the $F_n$. Thus $x$ must be a limit point of at least one of them, and thus, since they’re closed, an element.