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 $\epsilon < r$.
Best Answer
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.