Non-Hausdorff space such that all connected components are singletons

Question

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).

Answer 1

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.

Answer 2

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.