How can I prove that if $A$ is compact, then $A$ is finite? (Under the discrete metric) 
Let $\delta$ be the discrete metric on a non-empty set $X$.  Characterize the subsets of $X$ which are compact in $(X, \delta)$.

I remember the answer from a previous class: $A \subseteq X$ is compact $\iff A$ is finite.
I've proven one way in the general:

Suppose $S$ is a finite set.  Since there are only a finite
        number of terms, no sequence of distinct terms exists strictly
        in $S$, so $S$ does not have any limit points (and thus contains
        all of them; i.e. $\emptyset$).  Furthermore, since $S$ is
        finite, there exists a well-defined maximum $M = \max
  \{|s| : s \in S\}$.  Thus, $S$ is bounded by $M$. Since
        $S$ is closed and bounded, it is compact.

but I can't seem to go the other way.  How can I prove that, if $A \subset (X, \delta)$ is finite, then it is compact?

I had a look at Show that in a discrete metric space, every subset is both open and closed., but note that an infinite set can most certainly be bounded.

In a metric space, it is false in general that a bounded closed set is compact (for a counterexample, consider $\{q \mid 2< q^2<3\}$ in $\mathbb{Q}$).
You can prove that a finite set is always compact in a metric space using open coverings or subsequences. For the converse in discrete metric space, you can exhibit a sequence without converging subsequence in every infinite set (or exhibit an open covering whitout finite subcovering).
