Factor group of profinite group

Wikipedia (http://en.wikipedia.org/wiki/Profinite_group, Properties and Facts) says that the factor group of a profinite group $G$ by a closed normal subgroup $N$ is another profinite group. No proof or reference given and I'm trying to work this out for myself. Compactness and Hausdorffness seem easy enough but I don't know how to show that $G/N$ is totally disconnected. Thanks for any help.

You want by definition to prove that every connected components are singletons. This follows if you show that the connected component that contains the identity is a singleton. Let $C$ be this connected component containing $1$ in $G/N$. You want to show that $C= \{1\}$.
Let $$\pi : G \to G / N$$ be the projection map.
Now then, let $x\in G/N$, $x\neq 1$. You have the Hausdorff, so there is an open neighborhoos $U$ of $1$ such that $x\notin U$. So then $\pi^{-1}(U)$ is an open neighborhood of $1$ in $G$. Now $G$ is profinite, so $G$ has a neighborhood basis of $1$ consisting of compact open subgroups. Let $V$ be a compact open subgroup of $1$ in $G$ such that $V \subseteq \pi^{-1}(U)$.
Now $\pi(V)$ is open and compact in $G/N$. And $\pi(V)$ contains the connected component $C$. And $x \notin \pi(V)$.
So we have shown that given any $x\neq 1$ in $G/N$, $x$ is not contained in the connected component $C$ of $1$. So $C = \{1\}$.