Alexander's Compactness Theorem

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

Let $\mathcal B$ be a sub-basis of $\tau$.

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

Let the space $T$ have a sub-basis $\mathcal B$ such that every cover of $S$ by elements of $\mathcal B$ 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 $\mathcal C$ which has no finite subcover that is maximal among such open covers.

So if:
 * $V$ is an open set

and:
 * $V \notin \mathcal C$

then $\mathcal C \cup \left\{ {V}\right\}$ has a finite subcover, necessarily of the form:
 * $\mathcal C_0 \cup \left\{ {V}\right\}$

for some finite subset $\mathcal C_0$ of $\mathcal C$.

Consider $\mathcal C \cap \mathcal B$, that is, the sub-basic subset of $\mathcal C$.

Suppose $\mathcal C \cap \mathcal B$ covers $S$.

Then, by hypothesis, $\mathcal C \cap \mathcal B$ would have a finite subcover.

But $\mathcal C$ does not have a finite subcover.

So $\mathcal C \cap \mathcal B$ does not cover $S$.

Let $x \in S$ that is not covered by $\mathcal C \cap \mathcal B$.

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

We have that $\mathcal B$ is a sub-basis.

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

Since $x$ is not covered, $B_i \notin \mathcal C$.

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

But then $U$ and all the $\mathcal C_i$ cover $S$.

Hence $\mathcal C$ has a finite subcover.

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

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