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?
 A: You're a little muddled about what open and closed mean.  In a metric space, open and closed sets are defined using the concept of a ball.  This proof does not deal with metric spaces, but with topological spaces, which are more general, so there is no such thing as a ball here.  Here, to know whether the two halves of the separation are open, you just need to know whether they're in the topology $\tau$, and that's trivially true because any subset of the space is in $\tau$ by the definition of $\tau$.
A: 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.
A: I will only try to answer your question in update because I think that you have already gotten the answer of your other questions. If I am wrong, let me know.
This claim follows immediately if you assume the following as a definition of Connected Space.

Definition. A topological space $(X,\mathfrak{T})$ is said to be connected if the only clopen subsets of $X$ are $X$ and $\emptyset$.


If this is not the definition of connected space that you are using then we will use the following theorem,

Theorem. Let $(X,\mathfrak{T})$ be a topological space. Then $X$ is connected iff every continuous function $f:X\to\{\pm1\}$ is constant.

Let $X$ be a discreet topological space with at least two elements. If we are using this theorem then we need to find a continuous function $f:X\to\{\pm 1\}$ which is non-constant. For this purpose choose an $a\in X$ and define, $$f(x)=\begin{cases}1&\text{if}\ x=a\\-1&\text{if}\ x\ne a\end{cases}$$Can you show that the function is continuous (you will need to use the fact the in a topological space with discreet topology every closed set is also open)?

Reference
If you want to know about various equivalent statements of the definition of connected spaces that I wrote above see Theorem 2.3 of this.
