[Math] Closed subspace of a compact topological space is compact

Let $X$ be a compact topological space, and $A$ a closed subspace. Show that $A$ is compact.

How does this look?

Proof:

In order to show that $A$ is compact. We need to show that for any open cover $U$ of $A$, it contains a finite open cover of $A$.

Let $U$ be a open cover of $A$.

For each $U_\alpha \in U$, there exists an open set $V_\alpha \in X$ so that $V_\alpha \cap A = U_\alpha$.

Here we denote the index set of $U$ by $I$, and $\alpha \in I$.

Now, set $V = \{V_\alpha\}_{\alpha \in I}$, then $V \cap \{X \setminus U\}$ is an open cover of $X$.

It contains a finite subcover $\{V_1, …, V_n\} \cup \{X \setminus U\}$ of $X$,

hence $\{ V_1 \cap A, …, V_n \cap A \}$ is a finite subcover of $A$.

Hence $A$ is compact.