# Existence of a subsequence in compact metric space

Let $$\{x_n\}$$ be a sequence in a compact metric space X such that:

1) For all $$n \in \mathbb{N}$$ the set $$\{ x_n : n \ge N\}$$ is closed.
2) The sequence $$\{x_n\}$$ is Cauchy.

Prove that there exists $$x \in X$$ and a subsequence $$\{x_{n_k}\}$$ such that $$x_{n_k} =x$$ for all $$k \in \mathbb{N}$$.

Since we are in a compact metric space, the Cauchy sequence is convergent to some $$x \in X$$ and as $$\{x_n\}$$ is closed that would mean that $$x= x_M$$ for some $$m\in \mathbb{N}$$. But I'm not sure how to extract a subsequence which is eventually constant.

As you note, we can take $$x$$ to be the limit of the sequence $$x_n$$, which we know exists since compact spaces are complete.
In order to prove that the desired subsequence exist, it suffices to show that infinitely many elements of the sequence satisfy $$x_n = x$$. Suppose (for the purpose of contradiction) that only finitely many elements of the sequence satisfy $$x_n = x$$. Then there exists an $$N$$ such that the set $$\{x_n : n \geq N\}$$ does not contain the element $$x$$. The sequence $$\{x_n\}_{n \geq N}$$ converges to $$x$$, so $$x$$ is a limit point of the set $$\{x_n : n \geq N\}$$. The fact that $$x \notin \{x_n : n \geq N\}$$ contradicts the premise that $$\{x_n : n \geq N\}$$ is closed.