I had written out a full solution, but since this is homework, I've removed it and replaced it with this suggestion.
One way to prove that $T$ is totally disconnected is to show that whenever $p$ and $q$ are distinct points of $T$, then there are open sets $A$ and $B$ of $\mathbb{R}^2$ such that $T\subseteq A\cup B$, $(A\cap B)\cap T=\emptyset$, $p\in A$, and $q\in B$. This will show that there is a disconnection of $T$ in which $p$ and $q$ are in distinct components. In particular, $p$ and $q$ cannot be in the same connected component of $T$. If this holds for all pairs of points $p$ and $q$, then the connected components of $T$ must be single points.
So, pick two distinct points $p$ and $q$ in $T$. Try to find a line that is completely contained in $S$ and which separates $p$ and $q$. One way to achieve a line completely contained in $S$ is to have it go through a rational point with rational slope. Then throw away the line to get your two sets $A$ and $B$.
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
Hint: Find an open set $U$ containing the rationals whose measure is less than $1/2.$ Then look at $[0,1]\setminus U.$