H-Closed Space has a unique weaker topology which is Minimal Hausdorff

Question

Problem 17M of Willard is -

Every H-closed space has a unique weaker topology which is minimal Hausdorff.
(Hint: Use complements of regularly closed sets in $X$ as a base for a new topology)

I've been able to use the hint to construct a new topology, and have shown that it is a Hausdorff space. However, I've not been able to show that it is minimal Hausdorff, and that such a minimal Hausdorff space is unique. Any help would be appreciated!

Answer 1

Basic facts: Let $X$ be Hausdorff, then $X$ is $H$-closed iff every open filter on $X$ has a clusterpoint, and $X$ is $H$-minimal iff it is $H$-closed and semiregular (=the regular open sets form a base).

If $(X, \mathcal{T})$ is $H$-closed, and $\mathcal{T}'\subseteq \mathcal{T}$ is a Hausdorff topology, then $(X, \mathcal{T}')$ is $H$-closed (clear from the open ultrafilter characterisation). So the semiregularisation of $\mathcal{T}$ (which we can take as such a $\mathcal{T}'$, which is the topology suggested by Willard, is a Hausdorff seniregular and $H$-closed topology and hence $H$-minimal by known results.

So if $\mathcal{T}'$ is a $H$-minimal and coarser than $\mathcal{T}$ then we know it's $H$-closed and semiregular, and so it must be equal to the semiregularisation of $(X, \mathcal{T})$, showing unicity.