There are known the following implications regarding totally separated spaces:
- Every totally separated space is totally disconnected;
- Every totally separated space is Urysohn;
- Every zerodimensional and $T_0$ is totally separated space and
- Every extremally disconnected is totally separated space.
I need counterexamples for opposite implications i.e.:
- there is a totally disconnected space that is not totally separated;
- there is Urysohn space that is not totally separated;
- there is a totally separated space that is not zerodimensional or $T_0$ and
- there is a totally separated space that is not externally disconnected.
I suppose the book "Counterexamples in Topology" could help here.
Some hints:
Added: For 1. I got the following idea inspired by the counterexample 72 from the book. For every topological space $X$ we may form a “totally disconnected duplicated”: take $X × \{0, 1\}$, let $X × \{0\}$ be open discrete and let the basic neighborhood of $(x, 1)$ be $\{(x, 1)\} ∪ U × \{0\} \setminus \{(x, 0)\}$ for $U$ a neighborhood of $x$ in $X$. Denote this space by $X'$.
$X'$ is a duplicate of $X$ in the sense that $X'/\{(x, 0) \sim (x, 1)\}$ is canonically homeomorphic to $X$. $X'$ is scattered with two levels. If $X$ is $T_1$, then so is $X'$, and hence $X'$ is totally disconnected as any $T_1$ scattered space. On the other hand, the quasi-components of $X'$ look like this in the $T_1$ case: the singletons $\{(x, 0)\}$ are clopen quasi-components, while the quasi-components in $X × \{1\}$ correspond to quasi-components of $X$. In particular, if $X$ was connected and nondegenerate, than $X'$ is not totally separated. (I haven't checked everything so it may not be correct, it's an idea.)