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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.

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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.

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