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

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 $$\epsilon < r$$.

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.

• I see that in retrospect, but I always am pretty cautious when it comes to saying "directly implies" since I'm just starting to write longer proofs. I know that since the point $a$ is a limit point of $\{a_n\}$, it should be in $A$ since it's closed, but that is convergence in $X$. I was afraid it might not directly translate to convergence in $A$. I see that now, but for me it wasn't directly obvious so I tried to give another proof here :) – OneRaynyDay Nov 28 '18 at 6:30
• No worries! Keep up the good work! – JWP_HTX Nov 28 '18 at 6:48

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.