Subsequence and limits Any limit point of $\{$ $x_n$ $:$ $n\in\mathbb{N}$ $\}$ in a metric space $(X,d)$ is a subsequential limit of $(x_n)_n$
My attempt: Let $x$ be a limit point of $\{$ $x_n$ $:$ $n\in \mathbb{N}$ $\}$ =$A$. So, for each $r>0$, $B(x,r)\cap A\backslash \{$x$\}$ is non-empty. So in particular, for each $k\in \mathbb{N}$, pick $x_k\in B(x,\frac{1}{k}) \cap A\backslash \{$ x$\}$. This is clearly in $A$ and converges to $x$, by the squeeze principle. Now I need to show that this is a subsequence.
I am unsure as to how to proceed with this. Alternatively, one may construct a sequence inductively, in which case, I'd like to see the base case, induction hypothesis and etc setup, please.
 A: Choose $x_k\not\in\{x_1,\dots,x_{k-1}\}$ for each $k$.  This is possible unless $A$ is finite.
A: So let $(x_n)_n$ be a sequence in $(X,d)$, and let $A=\{x_n: n \in \Bbb N\}$ be its value set (which could be finite, or countably infinite). Suppose $x \in A'$.
Now, the set $B(x,1) \cap (A \setminus \{x\}$ is non-empty, so some $a_1$ is in it. This $a_1$ (being in $A$) is of the form $x_{n_1}$ for some $n_1 \in \Bbb N$. 
Going on by recursion: 
Suppose we're at stage $k$ and we already have indices $n_1 < n_2 < \ldots n_k$ such that for each $ i \le k$ we have $d(x,x_{n_i}) < \frac{1}{i}$.
Now define $r = \frac{1}{2}\min \left(\{d(x,x_n) : n \le n_k, x_n \neq x \} \cup \{ \frac{1}{k+1}\}\right) >0$
(mininum of finitely many positive numbers!) and pick $a \in B(x,r) \cap (A\setminus \{x\})$ and again this is of the form $a_n$ for some $n$. By the choice of $r$ we cannot have that $n \le n_k$ so $n_{k+1}:= n > n_k$ and also by the choice of $r$, $d(x_{n_{k+1}},x) < \frac{1}{k+1}$, so the recursion continues.
Now it's clear that $x_{n_k} \to x$ as $k \to \infty$: let $\varepsilon >0$ be given and find $K$ such that $\frac{1}{K} < \varepsilon$. Then for $k \ge K$ we have $d(x_{n_k}, x) < \frac{1}{k} \le \frac{1}{K} < \varepsilon$ showing convergence of this subsequence to $x$.
