$\textbf{Theorem}$: Let $(X,d)$ be a metric space and $A \subseteq X$. Then $(1)$ $A$ is sequentially compact if and only if $(2)$ $A$ is complete and completely bounded.
I am given a proof on this but there is a step that I don't quite understand. It concerns the proof of $(2)$:
$\textbf{$(1) \rightarrow (2)$}$: Suppose that $A$ is sequentially compact. Then $A$ is completely bounded (this was already proven in a previous section). Now suppose that $(x_{n})_{n}$ is a Cauchy sequence in $A$. Since $A$ is sequentially compact, there exists a subsequence $(x_{n_{k}})_{k}$ so that $x_{n_{k}} \rightarrow a$. Since $(x_{n})_{n}$ is a Cauchy sequence, then also $x_{n} \rightarrow a$. Thus $A$ is complete.
Now it is the last reasoning, namely that $x_{n} \rightarrow a$. Why will the Cauchy sequence converge to the same limit as its subsequence? I suppose it could be very obvious but alas I don't see it.