Is a countable, totally disconnected Hausdorff space necessarily totally separated? How about zero-dimensional? I'm starting to feel a little bad about using this website as my own personal counterexample generator, but here I go again... 
Terminology:
Let's call a space zero-dimensional if it is $T_0$ and admits a basis of clopen sets. By a standard embedding argument, a space is zero-dimensional if and only if it is homeomorphic to a subspace of some (possibly uncountable) power of the two-point discrete space $\{0,1\}$. In particular, zero-dimensional implies Hausdorff, or even completely regular.
Let's call a space $X$ totally separated if, given distinct points $x,y \in X$, there exists a separation $U,V$ of $X$ (ie. $U,V$ partition $X$ and are open) such that $x \in U$ and $y \in V$. In particular, a totally separated space is Hausdorff. Clearly zero-dimensional implies totally separated.
Finally, let's call a space totally disconnected if all its connected components are singletons. Clearly totally separated implies totally disconnected (conversely, totally disconnected need not even imply Hausdorff).
My question:
Let $X$ be a countable, totally disconnected Hausdorff space. Can $X$ fail to be totally separated? If yes, can $X$ fail to be zero-dimensional?
Some discussion:
If we replace "countable" with "compact, the answer to both questions is "no". For a compact Hausdorff space, the components and quasicomponents coincide, so $X$ is totally separated. Then, by applying basic compactness arguments, we can prove even that for all $A,B \subset X$, disjoint closed sets, there is a separation $U,V$ of $X$ with $A \subset U, B \subset V$. In particular, $X$ is zero-dimensional. The hypothesis of compactness cannot be dropped though. For example Cantor's leaky tent is a (noncompact, noncountable) subspace of the Euclidean plane which can be shown, with some effort, to be totally disconnected - but not totally separated. Since the hypothesis of compactness cannot be dropped, I wondered whether it could be replaced with something else. In particular, I wondered whether countable would do.
Added: Here's another counterexample. The main idea is the same as in Brian's example, but I thought this space seemed somehow more concrete.
As a set, let $X := \mathbb{Q} \cup \{p_0,p_1\}$ where $p_0,p_1$ are two distinct points not in $\mathbb{Q}$. For $n=0,1,2,\ldots$, let $I_n := (n,n+1) \cap \mathbb{Q}$. We put $U \subset X$ open if and only if the following are satisfied:


*

*$U \cap \mathbb{Q}$ is open in the standard topology on $\mathbb{Q}$.

*If $p_0 \in U$, then $U$ contains all but finitely many of $I_0,I_2,I_4,\ldots$

*If $p_1 \in U$, then $U$ contains all but finitely many of $I_1,I_3,I_5,\ldots$


It is easy to see this topology is Hausdorff. To see it is totally disconnected, suppose that $C \subset X$ is connected with more than one point. Intervals with irrational endpoints are still clopen, and these can be used to separate a fixed rational number from any other point in $X$. It follows that $C$ contains no rationals. Thus $C = \{p_0,q_0\}$, but this space is discrete ($X$ is Hausdorff), so no such $C$ exists. However, $X$ is not totally separated. Neighbourhoods of $p_0$ and $q_0$ cannot have disjoint closures.
 A: Let $X=(\omega\times\mathbb{Z})\cup\{p^-,p^+\}$, where $p^-$ and $p^+$ are distinct points not in $\omega\times\mathbb{Z}$. Let $Z_0=\mathbb{Z}\setminus\{0\}$. Points of $\omega\times Z_0$ are isolated. For each $n\in\omega$ and finite $F\subseteq Z_0$ let $$B(n,F)=\{n\}\times(Z_0\setminus F)\;,$$ and take $\{B(n,F):F\subseteq Z_0\text{ is finite}\}$ as a local base at $\langle n,0\rangle$. For $n\in\omega$ let $$B^+(n)=\{p^+\}\cup\{\langle i,k\rangle\in\omega\times Z_0:i>n\land k>0\}$$ and $$B^-(n)=\{p^-\}\cup\{\langle i,k\rangle\in\omega\times Z_0:i>n\land k<0\},$$ and take $\{B^+(n):n\in\omega\}$ and $\{B^-(n):n\in\omega\}$ as local bases at $p^+$ and $p^-$, respectively.
It’s easy to check that $X$ is Hausdorff and totally disconnected. However, $p^-$ and $p^+$ do not have open nbhds with disjoint closures: for any $n,m\in\omega$, $$\operatorname{cl}B^-(n)\cap\operatorname{cl}B^+(m)\supseteq\big\{\langle k,0\rangle:k>\max\{n,m\}\big\}\;.$$ It follows immediately that $X$ is neither totally separated nor zero-dimensional.
A: In Steen & Seebach, Counterexamples in Topology, p. 99, they prove that the Arens square is also an example of a countable, totally disconnected Hausdorff space that is neither totally separated nor zero dimensional.  I can post some details if nobody has the reference on hand.
A: Some vigorous googling turned up a paper containing an example of a countable, totally disconnected Hausdorff space which is not regular (hence not zero-dimensional). They use the example to demonstrate something rather more elaborate, so I have a feeling there is a lot of room to simplify the construction which I'll outline below.
As a set, we take $X$ to be the disjoint union of countably many copies of $\mathbb{N}$ (which I take to include $0$), one distinguished copy of $\mathbb{N}$, and an idealized point $p$.  So,
$$ X = (\mathbb{N}_0 \cup \mathbb{N}_1 \cup \mathbb{N}_2 \ldots ) \cup \mathbb{N} \cup \{p\}.$$
Now to topologize $X$. For this, we will need to fix an nonprincipal ultrafilter $\mathscr{P} \subset 2^\mathbb{N}$. For each $i$, let $\mathscr{P}_i$ denote the corresponding copy of $\mathscr{P}$.


*

*Each point in each $\mathbb{N}_i$ is taken to be isolated. 

*The neighbourhoods of $i \in \mathbb{N}$ are the sets that contain $\{i\} \cup P_i$ for some $P_i \in \mathscr{P}_i$.

*The neighbourhoods of $p$ are the sets that $\{p\} \cup \bigcup_{i \in P} \mathbb{N}_i$ for some $P \in \mathscr{P}$.


It is clear the open sets so defined are closed under unions. To show that the intersection of two neighbourhoods some $i \in \mathbb{N}$ or two neighbourhoods of $p$ is still a neighbourhood one just needs that filters are closed under finite intersections.
The topology we obtain is Hausdorff. For instance, we can separate $i \in \mathbb{N}$ from $x \in \mathbb{N}_i$ because $\mathscr{P}_i$ is nonprincipal so there is a $U_i \in \mathscr{P}_i$ with $x \notin U_i$ and $\{x\}$ and $\{i\} \cup U_i$ are the desired disjoint neighbourhoods.
The topology we obtain is totally disconnected. A connected subset $C$ of $X$ with 2 or more points cannot contain any points from the $\mathbb{N}_i$ since these points are open in $X$. Thus $C$ is a subspace of $Y := \mathbb{N} \cup \{p\}$. However, each point of $\mathbb{N}$ is open in $Y$, so none of these points can be in $C$ either and $C = \{p\}$ (contradiction).
Finally, the topology we obtain is not regular because we can't separate $p$ from the closed set $\mathbb{N}$. 
