Suppose $X$ is a zero dimensional Hausdorff space, and $\mathcal{B}$ a clopen basis for it.
Suppose $A$ is a set containing points $x$ and $y$. If they are distinct we can find an open set containing $x$ not containing $y$, hence we can find a basis element $B \in \mathcal{B}$ containing $x$ but not $y$. The sets $A\cap B$ and $A\cap B^c$ are both open in $A$ and disjoint, and their union is $A$, hence $A$ is disconnected.
Thus the space doesn't even need to be Hausdorff - the $T_1$ axiom is sufficient.
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
Let $(X, \tau) $ be totally disconnected topological space and $|X|<\infty$.
Since $X$ is totally disconnected, all components are singleton and components are closed implies every finite subsets of $X$ are closed.
Since $X$ is finite, every subsets of $X$ are closed implies $X$ is a discrete space.
Conclusion: A totally disconnected topological space where the underlying set is finite must be discrete.
Note: A finite $T_1$ space is always discrete.