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In rudin's analysis books, he defines compactness as: A subset $K$ of a metric space $X$ is $bf compact$ if every open cover of $K$ contains a finite subcover. More explicitly is that if $ G_alpha $ is an open cover of $K$ then one can select finitely many indices so that $K$ is contained in $G_alpha_1 cup ... cup G_alpha_n $

Rudin claims that a finite set is compact as an obvious fact. Im trying to verify this fact myself:

verification:

Let $F$ be a finite set and write it as $ a_1,...,a_n $ Take any open cover $ G_alpha $ of $F$. We can consider the open balls centered at $a_i$ and let $B_i$ be such ball. Then, $B_1 cup ... cup B_n$ contains $F$. Now, this seems incomplete as we must show that this finite union of open balls is in $cup_alpha G_alpha$. How do we check this?

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.

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.