In Theorem T000045 of pi-Base, a proof is given to defend the assertion from Counterexamples in Topology that all Extremally disconnected ($T_2$ where the closure of open is open) spaces are Totally separated (for every two points there is a disconnection of the entire space that separates them). A current pull request to pi-Base removes the $T_2$ requirement for the extremally disconnected property; however, does T45 still hold without such an assumption?
Extremally disconnected without Hausdorff
connectednessgeneral-topologyseparation-axioms
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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.
This notion of zero-dimensionality is based on the small inductive dimension $\operatorname{ind}(X)$ which can be defined for lots of spaces. It's defined inductively, based on the intuition that the boundary of open sets in two-dimensional spaces (which are boundaries disks, so circles in the plane, e.g.) have a dimension that is one lower than that of the space itself (this works nicely, at least intuitively, for the Euclidean spaces), setting up a recursion: We define $\operatorname{ind}(X) = -1$ iff $X=\emptyset$ (!) and a space has $\operatorname{ind}(X) \le n$ whenever $X$ has a base $\mathcal{B}$ of open sets such that $\operatorname{ind}(\partial O) \le n-1$ for all $O \in \mathcal{B}$, where $\partial A$ denotes the boundary of a set $A$. Finally, $\operatorname{ind}(X) = n$ holds if $\operatorname{ind}(X) \le n$ holds and $\operatorname{ind}(X)\le n-1$ does not hold. Also, $\operatorname{ind}(X)=\infty$ if for no $n$ we have $\operatorname{ind}(X) \le n$. It's clear that this is a topological invariant (homeomorphic spaces have the same dimension w.r.t. $\operatorname{ind}$) and zero-dimensional (i.e. $\operatorname{ind}(X)=0$) exactly means that there is a base of open sets with empty boundary (from the $-1$ clause!) and $\partial O=\emptyset$ means that $O$ is clopen.
Note that $\Bbb Q$ and $\Bbb P = \Bbb R\setminus \Bbb Q$ are both zero-dimensional in $\Bbb R$ and that $\Bbb R$ is one-dimensional (i.e. $\operatorname{ind}(\Bbb R)=1$) as boundaries of $(a,b)$ are $\{a,b\}$, which is zero-dimensional (discrete), etc.
In dimension theory more dimension functions have been defined as well, e.g. large inductive dimension $\operatorname{Ind}(X)$, which is a variant of $\operatorname{ind}(X)$, and the (Lebesgue) covering dimension $\dim(X)$, which has a different flavour and is about refinements of open covers and the order of those covers. For separable metric spaces however it can be shown that all $3$ that were mentioned are the same (i.e. give the same (integer) value). There are also metric-based definitions (fractal dimensions) which have more possible values, but are not topological, but metric invariants. Outside of metric spaces, we can have gaps between the dimension functions and stuff gets hairy quickly. See Engelking's book "Theory of dimensions, finite and infinite" for a taste of this field.
So in summary: the name "comes" (can be justified) from the small inductive dimension definition $\operatorname{ind}(X)$, but the name itself for that special class (clopen base) is older I think, and other names (like Boolean space etc) have been used too. It's a nice way to be very disconnected, giving a lot of structure.
Best Answer
You are right that the Hausdorff assumption is needed to show that extremally disconnected implies totally separated.
If we take extremally disconnected to mean the closure of any open set is open (without Hausdorff), then the following hold.
(1) Any hyperconnected space is extremally disconnected. (Reason: any nonempty open set $U\subseteq X$ is dense in $X$, so its closure is the whole space, which is open.)
(2) Any space that is at the same time extremally disconnected and connected is hyperconnected. (Reason: Any nonempty open set $U\subseteq X$ has a nonempty clopen closure. Since the space is connected, that closure must be the whole space, that is, $U$ is dense in $X$.)