Proof that a subspace $A$ of a complete metric space $X$ is complete iff $A$ is closed
Here's my proof in my own words, does it stack up?
Showing $A$ is complete implies $A$ is closed. Let $(x_n)$ be a convergent sequence in $A$. $A$ is complete $\implies (x_n) \to p \in A$. Hence $A$ is closed.
Showing $A$ is closed $\implies$ $A$ is complete. Let $(x_n)$ be a convergent sequence in $A$. $A$ is closed $\implies (x_n) \to p \in A$. As every convergent sequence is a Cauchy sequence, $(x_n)$ is a Cauchy sequence in $A$ that converges to $p \in A$ and hence $A$ is complete.
That sound ok?
As an aside, it seems to me that the definition of completeness and closed are basically identical, why are there two definitions for the same thing? Am I missing something here?
Your part 2 is slightly wrong. We need to show that $(x_n)$ $\in A$ Cauchy implies it converges to a limit in $A$. (This is different to what you have said, see my next paragraph) We do this by saying $X$ is complete, so $(x_n) \rightarrow x \in X$ and then use the fact that $A$ is closed in $X$ to say that since $(x_n) \rightarrow x \in X$, that $x\in A$.
This is not what you have said, since you mention that every convergent sequence is Cauchy (true) but it is not true in general that every Cauchy sequence is a convergent sequence (i.e. in incomplete metric spaces)
The fact that complete and closed are different concepts is because you can have closed sets in incomplete metric spaces. However, this theorem shows us that in any complete metric space, completeness and closure of a set is the same thing.