Alexander's Compactness Theorem

Proof
Let every open cover of $S$ have a finite subcover.

Let $\BB$ be a sub-basis of $\tau$.

By definition of a compact space, from every cover of $S$ by elements of $\BB$, a finite subcover can be selected.

Let the space $T$ have a sub-basis $\BB$ such that every cover of $S$ by elements of $\BB$ has a finite subcover.

$T$ is not such that every open cover of $S$ has a finite subcover.

Use Zorn's Lemma to find an open cover $\CC$ which has no finite subcover that is maximal among such open covers.

So if:
 * $V$ is an open set

and:
 * $V \notin \CC$

then $\CC \cup \set V$ has a finite subcover, necessarily of the form:
 * $\CC_0 \cup \set V$

for some finite subset $\CC_0$ of $\CC$.

Consider $\CC \cap \BB$, that is, the sub-basic subset of $\CC$.

Suppose $\CC \cap \BB$ covers $S$.

Then, by hypothesis, $\CC \cap \BB$ would have a finite subcover.

But $\CC$ does not have a finite subcover.

So $\CC \cap \BB$ does not cover $S$.

Let $x \in S$ that is not covered by $\CC \cap \BB$.

We have that $\CC$ covers $S$, so:
 * $\exists U \in \CC: x \in U$

We have that $\BB$ is a sub-basis.

So for some $B_1, \ldots, B_n \in \BB$, we have that:
 * $x \in B_1 \cap \cdots \cap B_n \subseteq U$

Since $x$ is not covered, $B_i \notin \CC$.

As noted above, this means that for each $i$, $B_i$ along with a finite subset $\CC_i$ of $\CC$, covers $S$.

But then $U$ and all the $\CC_i$ cover $S$.

Hence $\CC$ has a finite subcover.

This contradicts our supposition that we can construct $\CC$ so as to have no finite subcover.

It follows that we cannot construct an open cover $\CC$ of $S$ which has no finite subcover.

Also known as
Alexander's Compactness Theorem is also known as:
 * Alexander's Sub-Basis Theorem
 * Alexander's Sub-Base Theorem

which can also be seen in the form the Alexander Sub-Basis Theorem, and so on.

Sub-Base and Sub-Basis can also be seen here rendered as Subbase and Subbasis.