Closed Subspace of Compact Space is Compact

Theorem
A closed subspace of a compact space is compact.

That is, the property of being compact is weakly hereditary.

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 $$T \setminus C$$ 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.