I posted this answer here which a user pointed out to me is not correct. The question is asking for a proof that a compact metric space is complete.
My answer:
Note that in metric spaces the notions of compactness and sequential compactness coincide. Let $x_n$ be a Cauchy sequence in the metric space $X$. Since $X$ is sequentially compact there is a convergent subsequence $x_{n_k}\to x \in X$. But every subsequence of a convergent sequence converges to the same limit as the original sequence hence $x_n \to x \in X$. Hence $X$ is complete.
I believe this to be correct. By sequential compactness I obtain a convergent subsequence and since every subsequence of a Cauchy sequence converges to the same limit as the original sequence this is an argument that the original sequence converges.
Since I could not understand the comments by the other user I would like to kindly request the assistance of this commnuity to point out my error in different words.
You're confused. The fact for Cauchy sequences is: if $(x_n)$ is a Cauchy sequence (no assumptions on anything, like convergence!) and there exists some subsequence $(x_{n_k})$ of $(x_n)$ that converges to some $x \in X$, then $(x_n)$ converges to $x$ as well.
You say "it converges to the same limit as the original sequence", but you (on purpose!) do not assume at all that the original Cauchy sequence converges (this is what is to be shown!), so that statement makes no sense in this context. Correct is: the original sequence also converges to the same (subsequential) limit.
E.g. if we enumerate all rationals in a sequence, for every $x \in \mathbb{R}$ there exists some subsequential limit that converges to $x$. But this says nothing at all about the original sequence (which does not converge at all, and has all reals as subsequential limits). Of course this example sequence is not Cauchy...
So your proof works if we prove this first correct fact about Cauchy sequences and subsequential limits.