Let $(X,d)$ be a metric space. I can't find confirmation that $X$ is complete if and only if its closed subsets are exactly its complete subsets. Although it seems like it would be a useful thing to know. So I'm in doubt. Is this true ?
Here's my proof.
If $X$ is complete, then every Cauchy sequence of a closed subset $S$ converges. In metric spaces, closed subsets are sequentially closed hence the sequence converges in $S$ and $S$ is complete. If $S$, now, is assumed complete instead of closed, any of its convergent sequences is Cauchy, hence converges in $S$ by completeness, therefore $S$ is closed.
Now let's assume that $X$'s closed subsets are its complete subsets. Any Cauchy sequence is bounded, meaning it lies within a closed ball, therefore a complete ball, so it converges.
It is true simply because $X$ is a closed subspace of itself. Since we are assuming that the closed subsets of $X$ are exactly its complete subsets, it follows that $X$ is complete.
Your proof seems correct, though.