A Hausdorff topological space $(X,\mathcal T)$ is called H-closed or absolutely closed if it is closed in any Hausdorff space, which contains $X$ as a subspace.$\newcommand{\ol}[1]{\overline{#1}}\newcommand{\mc}[1]{\mathcal{#1}}$
A Hausdorff topological space $(X,\mathcal T)$ is called minimal Hausdorff if there is no Hausdorff topology on $X$ that is strictly weaker than $\mathcal T$. (I.e., it is minimal element of the set of all Hausdorff topologies on $X$ with partial ordering $\subseteq$.)
How to show that every minimal Hausdorff space is H-closed?
Is there an easy counterexample for the other direction? (I guess that finding a counterexample is not that easy - otherwise Willard would very probably mention an example.)
This is part of Problem 17M from Willard: General Topology, p.127
In the preceding problems from that book I have already shown some results on H-closed and minimal Hausdorff spaces:
Let $X$ be a Hausdorff space. $X$ is H-closed if and only if every open filter in $X$ has a cluster point.
Let $X$ be a Hausdorff space. $X$ is H-closed if and only if every open cover $\mc C$ of $X$ contains a finite subsystem $\mc D$ such that $\bigcup \{\ol D; D\in\mc D\}=X$, i.e., the closures of the sets from $\mc D$ cover $X$.
Let $X$ be a Hausdorff space. $X$ is minimal Hausdorff $\Leftrightarrow$ every open filter with unique cluster point converges.
Hint it Willard's book suggest to show that if $X$ Hausdorff and not H-closed, then it is not minimal. I've tried to use the characterization of H-closed spaces using filters. Which means that there is an open filter in $X$ which has no cluster point. But if I tried to use a construction similar to the one I used in the proof of the characterization of minimal Hausdorff spaces via open ultrafilters, the new topology was not necessarily strictly weaker than the original one.