If $X$ is a metric, then $X$ is compact if and only if $X$ is sequentially compact

Question

I'm going through a proof for the theorem:

If $X$ is a metric, then $X$ is compact if and only if $X$ is sequentially compact.

I'm trying to understand the easier forward direction but I'm having a hard time understanding where we've pulled/created a subsequence in this. So the proof goes like this:

Let $x_n$ be a sequence, and let $A_n = \{x_n, x_{n+1}, \ldots\}$ be a nested sequence of nonempty sets. Let $F_n = \overline{A_n}$ be nested and nonempty. By a corollary, the FIC holds, so $\bigcap_{n=1}^\infty F_n \neq \varnothing$.

Let $x \in \bigcap_{n=1}^\infty F_n$. Then $x \in F_1 = \overline{A_1}$, so $B(x,1) \cap A_1 \neq \varnothing$.

Then $\exists n_1 \geq 1$ such that $x_{n_1} \in B(x,1)$.

How is this last line deduced?

Answer 1

Since $x\in \overline{A_{1}}$ then the statement $B(x,1)\cap A_{1}\neq\emptyset$ follows from the definition of closure. But since $A_{1}=\{x_{1},x_{2},...\}$ then this can be rewritten as $B(x,1)\cap \{x_{1},x_{2},...\}\neq\emptyset$. Finally, by the definition of intersection we can conclude that there exists $n_{1}\geq 1$ such that $x_{n_{1}}\in B(x,1)$.