Showing that if a subset of a complete metric space is closed, it is also complete

To show this one must start with a Cauchy sequence, not a convergent sequence. Let $(x_n)$ be a Cauchy sequence in $A$. Then $(x_n)$ is a Cauchy sequence in $X$. Since $X$ is complete, $x_n\to x$ for some $x\in X$. Since $A$ is closed, $x\in A$. Hence $A$ is complete.

Let $(M,d)$ be a complete metric space, and let $A \subseteq M$. Suppose $A$ is closed.

Claim. $(A,d)$ is complete.

Proof. Let $(x_n)$ be a Cauchy sequence in $A$. Then $(x_n)$ is a Cauchy sequence in $(M,d)$ (trivial to verify). So $x_n \to x \in M$. But $A$ is closed, so it contains all of its limit points. So $x \in A$. Hence, $(A,d)$ is complete.

Here it is understood that the metric used on the subset say $N$ is the same as that used on $M$, the super set. Let $(x_n)$ be a sequence (Cauchy) in $N$. Regarding it as a sequence in $M$, it is still a Cauchy sequence in $M$ , and so since $M$ is complete, it converges in $M$ (say to $x$). Since, $x_n$ belongs to $N$ and $x_n \to x$ and $N$ is closed (all accumulation/cluster points are contained in $N$), $x$ belongs to $N$. Thus $(x_n)$ converges in $N$, and so $N$ is complete.