Compactness of finite sets

There is an open set that covers $a_1$

There is an open set that covers $a_2$

There is an open set that covers $a_3$

There is an open set that covers $a_4$

There is an open set that covers $a_5$

There is an open set that covers $a_6$

There is an open set that covers $a_7$

There is an open set that covers $a_8$

There is an open set that covers $a_9$

There is an open set that covers $a_{10}$

There is an open set that covers $a_{11}$

There is an open set that covers $a_{12}$

There is an open set that covers $a_{13}$

There is an open set that covers $a_{14}$

There is an open set that covers $a_{15}$

There is an open set that covers $a_{16}$

There is an open set that covers $a_{17}$

There is an open set that covers $a_{18}$

There is an open set that covers $a_{19}$

There is an open set that covers $a_{20}$

There is an open set that covers $a_{21}$ ... ... ...

There is an open set that covers $a_{n}$

It's a long but finite list, so it's a finite subcover.

There's no guarantee that any of the $B_i$ are in the set $\{G_{\alpha}\}$. For instance, We could have $\{G_{\alpha}\} = \{F\}$.

The point of compactness is that you really find a subcover of any given cover. In this case, we can argue as follows: suppose we are given a cover $\{G_\alpha\}$ of $F = \{ a_1, \ldots, a_n \}$. For all $i$, pick some $G_{\alpha_i}$ which contains $a_i$ (we know such a set exists because $\{G_\alpha\}$ covers $F$). Then $\{G_{\alpha_1},\ldots,G_{\alpha_n}\}$ is a finite subcover.

Notice how compactness really is a consequence of the finiteness of $F$, i.e. a property of $F$, and we aren't using any properties of the cover, since we don't know what it looks like a priori.