Is there an infinite, connected space $(X,\tau)$ such that the only Hausdorff topology $\tau_2$ with the property that $\tau\subseteq \tau_2$ is the discrete topology?

$\begingroup$ what if $X=\mathbb{N}$ and $\tau$ be the collection of sets of the form $\{n,n+1,...\}$? $\endgroup$ – erz Sep 21 '17 at 5:58

1$\begingroup$ @erz Seems to me any $\text T_1$ topology on $\mathbb N$ will be a refinement of your topology, e.g., a topology which is homeomorphic to the usual topology of $\mathbb Q,$ which is Hausdorff but not discrete. $\endgroup$ – bof Sep 21 '17 at 6:03
Remark. Every nondiscrete $\text T_1$ topology has a nondiscrete Hausdorff refinement.
Proof. Let $(X,\tau)$ be a nondiscrete $\text T_1$ space. Choose $p\in X$ such that $\{p\}\notin\tau,$ and let $\tau_2=\{A\cup B:A\in\tau,\ B\subseteq X\setminus\{p\}\}.$ Then $\tau_2$ is a Hausdorff topology on $X$ and $\tau\subseteq\tau_2$ and $\{p\}\notin\tau_2.$
Example. On any infinite set $X$ there is a connected $\text T_1$ topology $\tau$ such that no Hausdorff topology refining $\tau$ has more than one nonisolated point.
Namely, let $\tau=\mathcal U\cup\{\emptyset\}$ where $\mathcal U$ is a free ultrafilter on $X.$
Example. On any nonempty set $X$ there is a connected $\text T_0$ topology $\tau$ such that the only $\text T_1$ topology refining $\tau$ is the discrete topology.
Namely, choose $p\in X$ and let $\tau=\{A:p\in A\subseteq X\}\cup\{\emptyset\}.$