let $(X,d)$ be a metric space. How I can show that any finite subset of $X$ is 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.

For an answer that doesn't require any knowledge of separation axioms, consider the limit points of $\{ x \}$. Fix a point $y$ not in this set. Can $y$ be a limit point (hint: remember that $d(x,y) \ne 0$)?

Now remember that the finite union of closed sets is closed.

Metric spaces are hausdorff spaces (T2) and therefore T1. This means that points are closed and a finite union of closed sets, as is well known, is closed.