Totally separated, but the clopen sets do not form a basis

Letting $\mathcal{F}$ denote the family of all nonprincipal ultrafilters on $\omega$, set $X = \omega \cup \mathcal{F}$ with the topology obtained by

• taking each point of $\omega$ to be isolated; and
• for each $p \in \mathcal{F}$ taking the basic open neighbourhoods of $p$ to be of the form $A \cup \{ p \}$ where $A \in p$.

(This is called the strong ultrafilter topology in Steen and Seebach's Counterexamples in Topology.)

It can be shown that the clopen subsets of $X$ are exactly the subsets of the form $A \cup \{ p \in \mathcal{F} : A \in p \}$ for some $A \subseteq \omega$.

Using this it is straightforward to show that $X$ is totally separated. (E.g., if $p,q \in \mathcal{F}$ are distinct, take $A \in p \setminus q$.) It also follows that the clopen subsets of $X$ do not form a basis, since no clopen subset of $X$ can contain exactly one element of $\mathcal{F}$.

Let $X=\Bbb R\setminus\Bbb Q$. Fix an $\alpha\in X$, e.g., $\sqrt2$, and let $D=\alpha\Bbb Q=\{\alpha q:q\in\Bbb Q\}$. A set $U\subseteq X$ is open iff for each $x\in U$ there is an $\epsilon>0$ such that $(x-\epsilon,x+\epsilon)\cap D\subseteq U$. Equivalently, $D$ is a dense open subset of $X$ whose relative topology is its usual one, and each $x\in X\setminus D$ has a local base of open nbhds of the form $\{x\}\cup\big((a,b)\cap D\big)$ such that $a<x<b$. Let $\tau$ be the resulting topology.

Clearly $\tau$ is finer than the Euclidean topology, so $\langle X,\tau\rangle$ is totally separated. On the other hand, if $U=\{x\}\cup\big((a,b)\cap D\big)$ is a basic open nbhd of $x\in X\setminus D$, then

$$\operatorname{cl}_\tau U=U\cup\big([a,b]\setminus\Bbb Q\big)\nsubseteq U\;,$$

so $\langle X,\tau\rangle$ is not zero-dimensional.

If $X$ is the example due to Thomas that I described in this answer, $X\setminus\{p^-\}$ is another example. It’s clear from the construction that $Y$ is zero-dimensional, so $X\setminus\{p^-\}$ is totally separated. However, $\{p^+\}\cup\{\langle x,y\rangle\in Y:x>0\}$ is an open nbhd of $p^+$ that contains no clopen nbhd of $p^+$: if it did, the points $p^+$ and $p^-$ in $X$ could be separated by a continuous $\{0,1\}$-valued function. Thus, $X\setminus\{p^-\}$ is not zero-dimensional.

An easy way to get a totally separated space that is not $0$-dimensional is to start with a totally separated space and then enlarge the topology by adding a new open set, which will then typically not be a union of clopen sets. That is, start with a totally separated space $X$, pick a subset $A\subset X$, and let $Y$ be $X$ with the topology enlarged so that $A$ is open. So explicitly, an open subset of $Y$ is a set of the form $(A\cap U)\cup V$ where $U$ and $V$ are open in $X$. In particular, then, an open subset of $Y$ which contains the entire complement of $A$ must be open in $X$ (since in that case $(A\cap U)\cup V$ must actually be equal to $U\cup V$). Also, an open subset of $Y$ which is contained in $A$ has the form $A\cap U$ where $U$ is open in $X$. So, a clopen subset of $Y$ contained in $A$ must be closed in $X$ and must also have the form $A\cap U$ for $U$ open in $X$.

In particular, if $Y$ is to be zero-dimensional, then $A$ must be a union of clopen sets in $Y$. This means that for each $a\in A$, there is an open neighborhood $U$ of $a$ in $X$ such that $A\cap U$ is closed in $X$. But we can easily find an example of $X$ and $A$ such that this is not true: for instance, $X$ could be the Cantor set and $A$ could be a countable dense subset. So, in such an example, $Y$ will be totally separated but not $0$-dimensional.