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.
 A: 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).
A: First, in a metric space, a subset is compact if, and only if, every sequence of distinct elements in it has a convergent subsequence with limit in the subset. So, your argument for the first half can be slightly simplified (and in fact corrected, since the criterion you use for compactness is not quite right):there are no sequences of distinct elements in a finite subset, thus a finite subset is compact. 
Now, assume $A$ is compact in $X$. Suppose $A$ were infinite. Then there is some infinite sequence of distinct elements of $A$. By compactness, this sequence has a convergent subsequence with limit $L\in A$. But in the district metric, the only convergent sequences are the eventually constant ones. No subsequence of a set of distinct elements is eventually constant, thus a contradiction. 
Remark: the topological formulation of compactness (which is equivalent to the metric one (for metric spaces that is)), that a set is compact if, and only if, every open cover of it has a finite subcover gives the same result a lot quicker. If $A$ is compact, then cover it by its singletons. In the discrete metric, every singleton is open, so this is an open cover. By compactness, $A$ must be covered by a finite number of its singletons, so is itself finite. Conversely, if $A$ is finite and an open cover is given, then dilute the open cover to contain, for each element of $A$, just one open set containing that element. This gives a finite subcover. 
A: Hint Every $\{a\}$ is open hence 
$$A\subset\cup_{a\in A}\{a\}$$
is an open cover of $A$ and then there's finite subcover...
A: Hint: Show that a sequence in $(X, \delta)$ converges if and only if it is eventually constant.
A: Given any open cover of of any finite set, can be easily found a finite sub cover of the set. Then $A$ finite implies $A$ compact.
Now, in the discret metric no set have limit points. Then $A$ compact and $A$ infinite can't be hold, for if $E$ is an infinite subset of $A$ then $E$ must have a limit point in $A$.
