Proof that a discrete space (with more than 1 element) is not connected

Solution 1:

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 $\tau$.

Solution 2:

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.