Show that a finite subset of a metric space has no limit points and is therefore a closed set.
A subset of a metric space is closed if and only if it contains its limit points.
To be a limit point any neighborhood needs to contains infinitely many points within the set.
Thus, a finite subset cannot have any limit points.
Thus, the set of limit points is the empty set.
But, the finite subset contains the empty set.
Thus, the finite subset is closed.
This is an alternate solution to TPHs...it is a little more formal. Note, you may want to wait for confirmation of my solution by a more knowledgable member
Let $\displaystyle X$ be a metric space and let $\displaystyle \left\{p\right\}=A\subset X$. Now the only possibilities for a limit point of $\displaystyle A$ is the point $\displaystyle p$, but note that since there is only one point $\displaystyle A$ no matter how small you make an open ball around $\displaystyle p$ there will not be another point of $\displaystyle A$, thus a singleton in $\displaystyle X$ has no limit points, thus it is closed and consequently its complement is open. So now let another set $\displaystyle \left\{x_1,x_2,\cdots,x_n\right\}=B\subset X$ where $\displaystyle n$ is finite. It is clear that $\displaystyle B=\bigcup_{i=1}^{n} x_i$ we will now use the following lemma
Lemma: The intersection of any finite number of open sets is itself open.
Proof: Let $\displaystyle U_i~~1\leqslant i\leqslant n$ be open, then let $\displaystyle x\in\bigcap_{i=1}^{n}U_i$. This implies that $\displaystyle x\in U_i~\forall i$, and since each $\displaystyle U_i$ is open there exists an open ball $\displaystyle \mathcal{O}_{\varepsilon_i}(p,X)$ corresponding to each $\displaystyle sU_i$ such that $\displaystyle \mathcal{O}_{\varepsilon_i}(p,X)\subset U_i$. So now let $\displaystyle \varepsilon=\min\left\{\varepsilon_1,\varepsilon_2 ,\cdots\right\}$, then $\displaystyle \mathcal{O}_{\varepsilon}(x,X)\subset\bigcap_{i=1} ^{n}U_i\quad\blacksquare$
Using the above lemma and the fact that the complement of any singleton in $\displaystyle X$ is open we can see that the complement of any finite point set $\displaystyle \left[\bigcup_{i=1}^{n}x_i\right]^c=\bigcap_{i=1}^{n}x_i^c$ is open, thus the finite set is closed.