Proof that a discrete space (with more than 1 element) is not connected I'm reading this proof that says that a non-trivial discrete space is not connected. I understood that the proof works because it separated the discrete set into a singleton ${x}$ and its complementar. Since they're both open, their intersection is empty and their union is the entire space, this is a separation that is not trivial, therefore the space is not connected. But why a finite set of points is open? I remember that I proved that this set is closed, since I just have to pick a ball in the complementary, with radius such that its the minimum of the distances to those points. I know that if a set is closed it doesn't mean it's not open, but how to prove it?
Update: what's the simples proof that does not involve topology, only metrics?

It is true that finite sets are closed in every T$_1$ space, and thus they are closed in every discrete space.  Also by the definition of the discrete topology, $\textit{every}$ subset of the space is open. So suppose $X$ is discrete and has more than one point. Let $x\in X$. Then $\{x\}$ is open.  It is also closed (it is finite), and so its complement is also open (and nonempty). So $X$ is not connected. 
If you want to prove this in terms of metrics, the discrete topology on $X$ is induced by the metric $d(x,y)=0$ if $x=y$ and $d(x,y)=1$ if $x\neq y$. So if $x\in X$ then $$B_d (x,1)=\{y\in X:d(x,y)<1\}=\{x\}$$ and $$X\setminus \{x\}=\bigcup _{y\neq x} \{y\}=\bigcup _{y\neq x} B_d (y,0),$$ so $X$ is the union of two disjoint open sets.
