Question: Let $X$ be a discrete topology and $A$ is a subspace of $X$. Then,

$A$ is compact $\iff$ A is finite.

My attempt is general:

($\implies$) Let $X$ be compact. Suppose that X is infinite.

Consider a open-cover $C=\{\{x\} : x \in X\} $.

By the hypothesis, and by the definition of compactness of a space, 

$\exists H\in C : \bigcup\_{H\in C}H=X$ where H is finite.

Since $X$ is infinite but the subcover $H$ is finite, this is a contradiction.

Now, my question is: How can I solve this through a subspace $A$ of $X$?

Answer: 
You're proof is the right idea but you should be more formal about things when doing topology.
Assume that $A$ is compact and for the sake of contradiction, infinite. Then,
consider the following open cover of $A$, $C=\{\{x\}:x\in A\}$. Let $H$ be any finite open subcover, then we can write $H=\{\{x\_i\}:1\leq i \leq N\}$ for some $N\in\mathbb{N}$. Since $A$ is infinite, there is an element $y\in A$ such that $\{y\}\not\in H$. This contradicts that $H$ is a cover, so such a finite sub cover cannot exist.

For the other direction, assume that $A$ is finite. Then we have that $A = \bigcup\limits\_{i=1}^N \{x\_i\}$. Let $C$ be an open cover of $A$, $C=\{U\_j: j\in J\}$ for some open sets $U\_j$ with index set $J$ (possibly infinite). Since $C$ is a cover, each $x\_i\in A$ must belong to some open set in $C$, which we denote $U\_i$. Taking $K = \bigcup\limits\_{i=1}^N U\_i$ gives a finite open subcover of $A$, thus $A$ is compact (since $C$ was arbitrary, all open covers have a finite open subcover).
You don't have to. Compactness and discreteness are properties of spaces rather than of a subspaces. So if $A$ is a compact subset of a discrete space $X$, then $A$ is a compact discrete space on its own.

You should also note, that clearly a finite space is compact.