Possible Duplicate:
Compactness is closed-hereditary - significance of closed property?
Proposition let $(M,d)$ be a metric space where $K\subset M$ is compact and $A\subset K$ is closed. Then $A$ is compact.
Proof let $\{ G_\alpha \}_{\alpha\in I}$ be a open cover of $A$, then $ A \subset \cup_{\alpha\in I} G_\alpha $. Since $A^c$ is open and $K\setminus A \subset A^c$, $ K \subset \left( \bigcup_{\alpha\in I} G_\alpha \cup A^c\right) \quad (1) $ and, since $K$ is compact, $\exists N\in\mathbb{N}$, $\exists \alpha_1,...,\alpha_N \in I$: $ A\subset K \subset \left( \bigcup_{i\in I}^N G_{\alpha_i} \cup A^c\right) \quad (2) $ Then, $A$ is compact.
Question Why $A$ is necessarily closed? Step (1) is also true if $A$ is open since $\cup_{\alpha\in I} G_\alpha$ covers $A$, and $A\cup A^c = M$.
Thanks in advance.