I have to prove the fact that, given a metric space $(X,d)$ and a subset $K$ of $X$ compact, taking a closed subset $C$ of $K$, this $C$ is compact too.
I have used the characterization of closed sets with sequences, stating that there exist a generic sequence $Xn$ in $C$ that converges to an $x$ belonging to the same set $C$.
Then, using the fact that, since a sequence converges, every subsequence converges to the same limit, I stated that, under the Bolzano-Weierstrass property, set $C$ is compact since a generic sequence of its has a convergent subsequence to a some point of the set.
Now my question is: is that reasoning correct? Since in the solution of the question, the professor uses the fact that $K$ is compact to state the existence of a subsequence that converges.
Your argument is not correct. Every sequence in $C$ has a convergent subsequence because $K$ is compact. However, the subsequential limit is in $K$. You need to show that every subsequential limit is in $C$.
Let $\{x_n\}$ be a sequence in $C$. Because, $K$ is compact, $\{x_n\}$ has a convergent subsequence $\{x_{n_k}\}$ which converges to a point in $K$. Let $x=\lim_{k\to\infty}x_{n_k}$. So, $x\in\overline C$. But $C$ is closed, therefore $x\in C$.