I want to prove the following:
Let $X$ be a topological space. Remark: $x \sim y \iff$ There is a connected component which contains $x$ and $y$. And now I want to show that the quotient space $X/\sim $ is totally disconnected.
My way so far:
We suppose the opposite: $X/\sim $ is not totally disconnected. So there is a connected component $M\subset X/\sim $ which contains $2$ elements, named $[x]$ and $[y]$. $M$ is closed, hence $q^{-1}(M)$ is also closed in $X$ and contains $[x]$ and $[y]$. $[x]$ and $[y]$ are disjoint, hence $q^{-1}(M)$ is not connected, since it contains at least $2$ connected components. So there exists two open (not empty and disjoint) sets $U,V \subset$ $q^{-1}(M)$ with $U \cup V = q^{-1}(M)$.
But from now I don't know how to get a contradiction. Does someone have an idea to complete this proof?
Note that $q^{-1}(q(U))=U$, for if $u\in q^{-1}(q(U))$ then $q(u)=q(u')$ for some $u'\in U$. Hence $u$ and $u'$ live in the same connected component $C=q^{-1}(\{u\})\subseteq q^{-1}(M)$. Since $C=(C\cap U)\cup(C\cap V)$ and $C\cap U\neq\emptyset$, it follows $C\cap V=\emptyset$, thus $C\subseteq U$.
Consequently $M=q(U)\cup q(V)$ and $q(U)$, $q(V)$ are nonempty, open, disjoint subsets of $M$.