Closed Subspace of Compact Space is Compact

Theorem
A closed subspace of a compact space is compact.

Proof
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 $$T - C$$ is open in $$T$$.

So if we add $$T - C$$ to $$\mathcal{U}$$, we get 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 also covers $$C$$ by the fact that it covers $$T$$.

If $$T - C$$ 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.