Non-Hausdorff space such that all connected components are singletons Is there a topological space $(X,\tau)$ such that


*

*$(X,\tau)$ is not Hausdorff;

*if $S\subseteq X$ and $S$ contains more than 1 point, then $S$ is not connected (with the subspace topology).

 A: Yes. Let $X=\Bbb N\cup\{p,q\}$, where $p$ and $q$ are distinct points not in $\Bbb N$. Let $\tau$ be the topology generated by the base
$$\begin{align*}
\mathscr{B}=\big\{\{n\}:n\in\Bbb N\big\}&\cup\big\{\{p\}\cup(\Bbb N\setminus F):F\subseteq\Bbb N\text{ is finite}\big\}\\
&\cup\big\{\{q\}\cup(\Bbb N\setminus F):F\subseteq\Bbb N\text{ is finite}\big\}\;;
\end{align*}$$
in effect $X$ is a simple sequence converging to two distinct limits, $p$ and $q$. Clearly $X$ is not Hausdorff, since $p$ and $q$ cannot be separated by disjoint open sets. Equally clearly $X$ is totally disconnected.
A: Let $X$ be the space
$$\{0,0'\}\cup\{1/n\mid n\in\Bbb N\}$$
where the neighborhoods of a point $1/n$ are all subsets of $X$ containing $1/n$, and $U$ is a neighborhood of $0$ or $0'$ iff $U$ contains that point and almost all $1/n$.
You can also think of this space as a quotient of
$$\{(0,0),(0,1)\}\cup\bigcup_{n\in\Bbb N}\{1/n\}\times I$$
with the subspace topology of $\Bbb R^2$, obtained by identifying each connected component to a point.
This space is not Hausdorff but every subset is disconnected.
