Compact Subspace of Hausdorff Space is Closed

Theorem
Let $$H = \left({A, \vartheta}\right)$$ be a Hausdorff space.

Let $$C$$ be a subspace of $$H$$.

If $$C$$ is compact, then it is closed.

Corollary
Singleton point sets and finite subsets of a Hausdorff space are closed.

Proof
From Basic Properties of a Hausdorff Space, any subspace of a Hausdorff space is itself Hausdorff.

Let $$a \in H - C$$.

We are going to prove that there exists an open set $$U_a$$ such that $$a \in U_a \subseteq H - C$$.

For any single point $$x \in C$$, the Hausdorff condition ensures the existence of disjoint open set $$U \left({x}\right)$$ and $$V \left({x}\right)$$ containing $$a$$ and $$x$$ respectively.

Suppose there were only a finite number of points $$x_1, x_2, \ldots, x_r$$ in $$C$$.

Then we could take $$U_a = \bigcap_{i=1}^r U \left({x_i}\right)$$ and get $$a \in U_a \subseteq H - C$$.

Now suppose $$C$$ is not finite.

The set $$\left\{{V \left({x}\right): x \in C}\right\}$$ is an open cover of $$C$$.

As $$C$$ is compact, it has a finite subcover, say $$\left\{{V \left({x_1}\right), V \left({x_2}\right), \ldots, V \left({x_r}\right)}\right\}$$.

Let $$U_a = \bigcap_{i=1}^r U \left({x_i}\right)$$.

Then $$U_a$$ is open because it is a finite intersection of open sets.

Also, $$a \in U_a$$ because $$a \in U \left({x_i}\right)$$ for each $$i = 1, 2, \ldots, r$$.

Finally, if $$b \in U_a$$ then for any $$i = 1, 2, \ldots, r$$ we have $$b \in U \left({x_i}\right)$$.

Hence $$b \notin V \left({x_i}\right)$$, so $$b \notin C$$, since $$C = \bigcup_{i=1}^r V \left({x_i}\right)$$.

Thus $$U_a \subseteq H - C$$.

Then $$H - C = \bigcup_{a \in H - C} U_a$$, so $$H - C$$ is open and so $$C$$ is closed.

Proof of Corollary
Follows directly from the above.