Lef $(X,\tau)$ a topological space: Show that if $\{A_n: n \in \mathbb{N}\}$ is a family of subsets of $ X $ such that $ A_{n+1} \subset A_n $ $ \forall n \in \mathbb{N}$ , y $\bigcap_{n \in \mathbb{N}} \ \overline{A_n} =\emptyset$, then $\{A_n: \ n \in \mathbb{N}\}$ is locally finite.
Seeing that $\{A_n : n \in \mathbb{N}\} $ is locally finite, is equivalent to seeing that given a point $x \in X$ there exists a neighborhood $ U $ such that the set $\{n \in \mathbb{N}: U \cap A_n \neq \emptyset\}$ is finite.
For this I remembered the property of Cantor: Given any countable family of closed $\{F_n \ : n \in \mathbb{N}\}$, non-empty and fitted sets ${F_n}$ ($F_ {n + 1} \subset F_n$) then $ \bigcap_{n \in \mathbb{N}} F_n \neq \emptyset$, but in the hypothesis it says that this does not happen.
The Cantor Intersection Theorem is for decreasing nests of compact sets, so it doesn’t apply here. We don’t need anything that fancy anyway. Let $x\in X$. Then $x\notin\bigcap_{n\in\Bbb N}\operatorname{cl}A_n$, so there is an $n\in\Bbb N$ such that $x\notin\operatorname{cl}A_n$, and therefore $x$ has an open nbhd $V$ such that $V\cap A_n=\varnothing$. Now use the fact that the sets $A_k$ are a decreasing nest to conclude that $\{k\in\Bbb N:V\cap A_k\ne\varnothing\}$ is finite.