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 $$S - \left\{{x}\right\}$$ 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 $$x_{n \left({1}\right)} \in N_1 \left({x}\right)$$, where $$N_1 \left({x}\right)$$ is the $1$-neighborhood of $$x$$.

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

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

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 $$x_{n \left({k+1}\right)} \in N_{\epsilon} \left({x}\right)$$.

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 $$x_{n \left({k+1}\right)} \in N_{\frac 1 {k+1}} \left({x}\right)$$.

This completes the inductive step in constructing a subsequence.

This converges to $$x$$, because $$\forall k_0 \in \N^*$$ and $$\forall k \ge k_0$$ we have $$d \left({x_{n \left({k}\right)}, x}\right) < \frac 1 k \ge \frac 1 {k_0}$$.

Hence the result.