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 $M$.
This is too long to be a comment and is currently only a partial answer.
Following the idea that the existence of an open, totally disconnected set should relate to the connectedness of the space, I read about ways a space can be connected. A locally connected space is one where every point in a open set has a connected, open neighborhood and felt most relevant. I wasn't able to show that being locally connected was a sufficient condition for the lack of an open, totally disconnected set. So, I ended up adding connected back as a requirement (and T1 ended up being helpful during the proof) which led to,
Let $(X,\tau)$ be a locally connected, connected, T1 space that is not the topology on a one point set. Then, $(X, \tau)$ is not weakly disconnected.
Proof:
The proof strategy will be to use proof by contrapositive. Let $(X, \tau)$ be a weakly disconnected space. Let $U$ be an open, non-empty totally disconnected set. Let $(U, \tau_{U})$ be the subspace topology of $U$. As $U$ is totally disconnected and non-empty, its connected components are singletons.
Here, we'll split the proof into two cases. Either all of its connected components are open or they there exists a connected component that is not open.
If there exists a connected component that is not open then its connected components are not all open which means it is not locally connected. As locally connected is hereditary for open subspaces, $(X,\tau)$, can not be locally connected.
If there exists a connected component that is open, call that component $A$. As $U$ is totally disconnected, $A$ must be a singleton. Since $U$ is open, any open set in $U$ is also open in $X$. This means $A$ is an open singleton in $X$. As $X$ is a T1 space, singletons are closed making $A$ a clopen set. Since $X$ is not a one point set, $A$ is non-trivial clopen set. This means $(X,\tau)$ is not connected.
In either case, $(X, \tau)$ can not have the three properties of being locally connected, connected, and T1. This completes the proof.
The issue is I'm not sure if those three properties are a necessary condition, so I still lack a nice way of describing all spaces that lack an open, totally disconnected set. While my current proof strategy uses the fact the space is T1, I'm skeptical that there isn't an alternate way to drop that condition.
Best Answer
Yes, it’s true.
Let $X=\prod_{\alpha\in A}X_\alpha$, where each $X_\alpha$ is totally disconnected. To show that $X$ is totally disconnected, you just have to show that if $x$ and $y$ are distinct points of $X$, then there is a clopen set $U$ such that $x\in U$ and $y\notin U$: that’s exactly what it means to say that $x$ and $y$ are in different components of $X$.
Suppose, then, that $x=\langle x_\alpha:\alpha\in A\rangle,y=\langle y_\alpha:\alpha\in A\rangle\in X$, and $x\ne y$. Then there is some $\alpha\in A$ such that $x_\alpha\ne y_\alpha$. $X_\alpha$ is totally disconnected, so there is a clopen set $U$ in $X_\alpha$ such that $x_\alpha\in U$ and $y_\alpha\notin U$. Let $V=\pi_\alpha^{-1}[U]$, where $\pi_\alpha:X\to X_\alpha$ is the usual projection map; then $V$ is a clopen set in $X$, $x\in V$, and $y\notin V$.