The discrete metric just says that
$$d(x,x)=0$$
$$d(x,y)=1,\ x\neq y$$
So say your ball has radius $r$. If $r<1$ then the only point it contains is the point it's centred on. So any single point has a ball of some radius around it containing only that point. This is the same thing as $B_{0<r<1}(x)=\{x\}$, so we know that every singleton is open. And now we're actually done! Since now we know that any point $x$ in a set $A$ has a ball containing it, because we can always construct a ball that only contains $x$! Since all sets are open, their complements are open as well. This implies that all sets are also closed.
For an approach even more basic than Andrew Salmon’s, let $\langle X,d\rangle$ be a metric space, and let $F$ be any finite subset of $X$. The empty set is closed by definition, so we might as well assume that $F\ne\varnothing$. Now suppose that $x\in X\setminus F$, and let $r_x=\min\{d(x,y):y\in F\}$. Then $r_x>0$ (why?); what can you say about $B(x,r_x)$, the open ball of radius $r_x$ centred at $x$?
Yes, a finite set in a metric space can be open. First, the empty set is always open. Other than that, though, it depends on the space. No finite, non-empty subset of $\Bbb R^n$ is open, for instance, for any $n\in\Bbb Z^+$. However, if $X$ is any set at all, the function $d:X\times X\to\Bbb R$ defined by
$$d(x,y)=\begin{cases}1,&\text{if }x\ne y\\0,&\text{if }x=y\end{cases}$$
is a metric, often called the discrete metric, and every subset of $X$ is open.
Best Answer
Hint: If $(A,d)$ is a finite metric space and $x \in A$ and we let $$\delta=\min_{y \in A \setminus \{x\}}d(x,y)$$ then what is in $B(x,\delta)$?