Let (X, d) be a metric space.
Show that any finite subset is compact.
Are you serious? Let $\displaystyle \mathcal{V}$ be an open cover for some finite subset $\displaystyle \mathcal{S}$ of $\displaystyle X$. There must exists some $\displaystyle O_s\in\mathcal{V}$ such that $\displaystyle s\in O_s$ for each $\displaystyle s\in\mathcal{S}$. Clearly then $\displaystyle \left\{O_s\right\}_{s\in\mathcal{S}}$ is a subcover of $\displaystyle \mathcal{S}$ and since $\displaystyle \mathcal{S}$ is finite so is $\displaystyle \left\{O_s\right\}_{s\in\mathcal{S}}$.
While , what h2osprey said was undeniably correct, I think the finer points got lost.
To be compact means that every open cover has a finite subcover. Thus, to show that a set is compact one way is to take an arbitrary open cover and derive a finite subcover. So, the concept is we let $\displaystyle \mathcal{V}$ be some open cover of $\displaystyle \mathcal{S}$. Next, we let $\displaystyle s\in\mathcal{S}$ be arbitrary, and since $\displaystyle \mathcal{V}$ covers $\displaystyle \mathcal{S}$ there exists some open set, which we will call $\displaystyle O_s$, in $\displaystyle \mathcal{V}$ such that it contains $\displaystyle s$. So, for each $\displaystyle s\in\mathcal{S}$ we take some corresponding $\displaystyle O_s\in\mathcal{V}$ and construct a new class $\displaystyle \left\{O_s\right\}_{s\in\mathcal{S}}$. This is clearly a subcover, and since there is at most $\displaystyle \text{card }\mathcal{S}$ elements of $\displaystyle \left\{O_s\right\}_{s\in\mathcal{S}}$ it must be a finite subcover.
The two finer points that you should definitely answer yourself is why can I choose this $\displaystyle O_s$ and why is there at most $\displaystyle \text{card }\mathcal{S}$ elements in $\displaystyle \left\{O_s\right\}_{s\in\mathcal{S}}$