How can I prove that if $A$ is compact, then $A$ is finite? (Under the discrete metric)

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.

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...