Prove a closed subset of a complete metric space is complete via contradiction

Question

I've seen some proofs by definition and just want to ask for proof verification on whether this is okay as well:

Given complete metric space $(X,d)$ and $A \subset X$, $A$ closed. Prove $A$ is complete.

We know any cauchy sequence in $X$ converges, i.e. $\{a_n\} \subset X$ and cauchy implies $a_n \to a \in X$.

Suppose for contradiction, $A$ is not complete, then $\exists \{a_n\} \subset A$ and cauchy such that $a_n$ does not converge in A. Since $A \subset X$, this sequence converges to $a \in X/A$.

Since $A$ closed, $X/A$ open, so $\exists r > 0$ s.t. $B_r(a) \subset X/A$. Then this contradicts that it is cauchy by picking any $\epsilon < r$.

Answer 1

This is basically fine, but you don't really need to go by contradiction. $A\subset X$ is closed iff it contains all of its limit points. Just let an arbitrary Cauchy sequence, $\{a_{n}\}\subset A$ be given, it must converge in $X$ by completeness, and then the closedness directly implies that this limit is in $A$, making $A$ a complete metric space in its own right.

Answer 2

Sure, this proof works!

Just a few comments, though: firstly, take the sentence

We know any cauchy sequence in $X$ converges, . . .

In my opinion, it would be preferable to say that

We know any Cauchy sequence in $X$ converges in $X$, . . .

just for the sake of more clarity, even though you have explained what this means by adding an "i.e.".

Secondly,

Then this contradicts that it is cauchy by picking any $\epsilon < r$.

It will be helpful to explain how the contradiction arises. At first glance, I can see that there will be a contradiction to the fact that $\{ a_n \}$ converges to $a$, not that $\{ a_n \}$ is Cauchy. But, the two ideas are closely related enough that I am still confident that you're not necessarily wrong. So, adding more details here would be better, in my opinion.