Closed Subspace of Compact Space is Compact

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Theorem

Let $T$ be a compact space.

Let $C$ be a closed subspace of $T$.

Let $\mathcal U$ be an open cover of $C$.

Since $C$ is closed, it follows by definition of closed that $T \setminus C$ is open in $T$.

So if we add $T \setminus C$ to $\mathcal U$, we see that $\mathcal U \cup \left({T \setminus C}\right)$ is also an open cover of $T$.

As $T$ is compact, there is a finite subcover of $\mathcal U$, say $\mathcal V = \left\{{U_1, U_2, \ldots, U_r}\right\}$.

This covers $C$ by the fact that it covers $T$.

If $T \setminus C$ is an element of $\mathcal V$, then it can be removed from $\mathcal V$ and the rest of $\mathcal V$ still covers $C$.

Thus we have a finite subcover of $\mathcal U$ which covers $C$, and hence $C$ is compact.

$\blacksquare$