# If (X,d) is a metric space and $K \subseteq X$ is compact. Show that $K$ is complete.

I've seen other proofs of this on here but all of them seem to rely on showing compact iff sequentially compact in a metric space but instead I want to use only the definition of compactness, that every open cover of a compact space admits a finite subcover to prove this.

Suppose K is compact and not complete. Then there is a Cauchy sequence in K that is not convergent in K. Let $$(x_n)$$ be this sequence. I want to somehow show that this will give an open cover with no finite subcover but I can't see how I do that. Or I want to show that somehow the cluster point for the cauchy sequence must be in K. But I can't see how I can show that.
It is not hard to prove that if a Cauchy sequence has a convergent subsequence, then the whole sequence converges. So, suppose that $$(x_n)_{n\in\mathbb N}$$ is a Cauchy sequence that doesn't converge. If $$x\in K$$, then $$x$$ is not the limit of a a subsequence of $$(x_n)_{n\in\mathbb N}$$. So, there is an open set $$A_x$$ such that $$x\in A_x$$ and such that $$A_x$$ contains only finitely many terms of the sequence. Note that $$K=\bigcup_{x\in K}A_x$$; so, $$(A_x)_{x\in K}$$ is an open cover of $$K$$. Therefore, it hs a finite subcover $$(A_{x_j})_{j\in\{1,2,\ldots,N\}}$$. But since each $$A_{x_j}$$ contains finitely many $$x_n$$'s, this is impossible. A contradiction has been reached and so the sequence $$(x_n)_{n\in\mathbb N}$$ converges.