Let $(A_{j})$ be a sequence of closed subsets of $X$ with $A_{j} \supseteq A_{j+1}$ for all $j \in \mathbb{N}$, then $\cap A_{j} \neq \emptyset .$ Good afternoon, I'm doing Problem III.3.5 from textbook Analysis I by Amann.


Here the underlying space is metric space.

Could you please verify if my proof looks fine or contains logical gaps/errors? Any suggestion is greatly appreciated.
My attempt:
By Axiom of Countable Choice, there is a sequence $(x_n)$ in $A_0$ such that $x_n \in A_n$ for all $n$. Since $A_0$ is compact, there is a subsequence $(x_{n_m})$ of $(x_n)$ such that $(x_{n_m})$ converges to $l \in A_0$. Because $x_{n_m} \in A_{n_m}$ and $A_{n_m}$ is closed for all $m$, then $$\forall m:l =  \lim_{m \to \infty} (x_{n_m}) \in A_{n_m}$$ and consequently $$l \in \bigcap_{m=0}^\infty A_{n_m} = \bigcap_{n=0}^\infty A_{n}$$ As such, $\bigcap_{n=0}^\infty A_{n} \neq \emptyset$.
 A: A small refinement to show that $l \in \bigcap A_n$: Recall that we have a nested sequence of closed sets $A_n$ with $x_n \in A_n$ for all $n$ and $l \in X$ and a  subsequence $x_{n_k} \to l, (k \to \infty)$.
Fix $n_0 \in \mathbb{N}$. Then for some $k_1 \in \mathbb{N}$ we have that $n_{k_1} \ge n_0$ (subsequences are cofinal). Then for all $k \ge k_1$ we have $$x_{n_k} \in A_{n_k} \subseteq A_{n_{k_1}} \subseteq A_{n_0}$$ using the nestedness of the sets.
So $$l=\lim_{k \to \infty} x_{n_k}= \lim_{k \ge k_1, k \to \infty} x_{n_k}$$
as the limit of a sequence only depends on the tail. And as all the terms in the last limit are in the closed set $A_{n_0}$ as we saw, $l \in A_{n_0}$. As $n_0$ was arbitrary, $l \in \bigcap_n A_n$ which is thus non-empty.  
A: The statement of Problem III.3.5 is valid for any space $X,$ but the general case is not provable by use of sequences, as that method is inapplicable to many non-metrizable spaces. 
Definition. A space $Y$ is compact iff every open cover of $Y$ has a finite sub-cover.
An equivalent def'n, involving closed sets instead of open sets, is: A space $Y$ is compact iff $\cap F\ne \emptyset$ whenever $F$ is a non-empty family of closed sets such that $F$ has the FIP (Finite Intersection Property).
"$F$ has the FIP" means that $\cap G \ne \emptyset$ whenever $G$ is a finite non-empty subset of $F.$
So let $Y=A_0.$  Each $Cl_X(A_n)=A_n\subset A_0=Cl_X(A_0),$ so each $A_n$ is closed in the $space$ $Y.$ And since $A_n\supset A_{n+1}\ne \emptyset,$ the family $F=\{A_n: n\ge 0\}$ has the FIP. Now since $Y$ is compact, therefore $\cap F\ne \emptyset.$
