Subsequence of Sequence in Metric Space with Limit

Theorem
Let $M = \left({A, d}\right)$ be a metric space.

Let $\left \langle {x_n} \right \rangle$ be a sequence in $M$.

Let $x$ be a limit point of $S = \left\{{x_n: n \in \N}\right\}$, the set of members of $\left \langle {x_n} \right \rangle$.

Then $\left \langle {x_n} \right \rangle$ has a subsequence which converges to $x$.

Proof
By Finite Subset of Metric Space has no Limit Points‎, $S$ is infinite (or it has no limit points).

We may assume that $x_n$ never equals $x$.

Otherwise we delete all instances of $x$ from $\left \langle {x_n} \right \rangle$ and create a new sequence $x_m$ which is a subsequence of $x_n$ which does never equal $x$.

Then $S \setminus \left\{{x}\right\}$ is still infinite, and it still has $x$ as a limit point.

So, since $x$ is a limit point of $S$, there is an integer $n \left({1}\right)$, say, such that $x_{n \left({1}\right)} \in B_1 \left({x}\right)$, where $B_1 \left({x}\right)$ is the open $1$-ball of $x$.

Suppose we choose the integers $n \left({1}\right) < n \left({2}\right) < \cdots < n \left({k}\right)$ so that $x_{n \left({i}\right)} \in B_{1/i} \left({x}\right)$ for $i = 1, 2, \ldots, k$.

Now we put $\epsilon = \min \left\{{\dfrac 1 {k+1}, d \left({x_1, x}\right), d \left({x_2, x}\right), d \left({x_{n \left({k}\right)}, x}\right)}\right\}$.

Since $x_n \ne x$ for any $n$, it follows that $\epsilon > 0$.

Since $x$ is a limit point of $S$, there exists an integer $n \left({k+1}\right)$ such that $x_{n \left({k+1}\right)} \in B_\epsilon \left({x}\right)$.

Now $\epsilon$ has been chosen so as to force $n \left({k+1}\right) > n \left({k}\right)$, since $d \left({x_i, x}\right) \ge \epsilon$ for all $i \le n \left({k}\right)$.

Also, note that $x_{n \left({k+1}\right)} \in B_{\frac 1 {k+1}} \left({x}\right)$.

This completes the inductive step in constructing a subsequence.

This converges to $x$, because $\forall k_0 \in \N_{>0}$ and $\forall k \ge k_0$ we have $d \left({x_{n \left({k}\right)}, x}\right) < \dfrac 1 k \ge \dfrac 1 {k_0}$.

Hence the result.