a question about finer topology A Hausdorff topological space $(X,\mathcal T)$ is called ‎‎$‎H‎$‎-closed   if it is closed in any Hausdorff space, which contains $X$ as a subspace.
I ‎want ‎to ‎prove ‎that ‎‎‎the topological property ‎of‎ ‎$‎H‎$‎-closed is established for finer topologies, in other words ‎‎$‎(X, ‎\tau_2)‎$ ‎is ‎‎$‎H‎$‎- closed‎‎ and $‎‎ \tau_1‎‎\subset‎  ‎\tau_2‎$. But I do not know if my method is correct or not.
I assume ‎‎$‎(X ,‎\tau_1)‎$‎‎  is a ‎‎$‎H‎$‎- closed space ‎and‎  $‎‎\tau_1‎‎\subset‎  ‎\tau_2‎$.‎
‎
According to the definition,  space ‎‎$‎(X,‎\tau_2)‎$‎‎ can be ‎imbedded‎ in a Hausdorff space $ ‎‎(Y, ‎\tau)‎‎‎‎‎‎‎$‎, t‎hat ‎is, ‎‎$‎‎  ‎\tau_2 = \tau_‎\mid_{X}  ‎‎‎$‎.‎‎
How can I ‎use from ‎$‎H‎$‎-clossed ‎property ‎of ‎$‎(X ,‎\tau_1)‎$   ‎ ‎to say that ‎there is ‎a‎ ‎Hausdorff‎ topology ‎‎$‎\tau‎^{‎*‎}‎‎‎$  so ‎that‎ ‎$‎‎\tau‎^{*} ‎\subset ‎\tau‎‎‎‎$ and ‎$‎‎\tau‎^{‎*‎}‎_‎\mid_X = ‎\tau_1‎$‎‎‎?‎‎
 A: Thoughts:
In the situation where $(X,\tau_1)$ is (Hausdorff and) $H$-closed, and $(X,\tau_2)$ would be $H$-closed and we have a proper inclusion $\tau_1 \subsetneq \tau_2$, then $(X,\tau_2)$ cannot be semiregular, by the theorem

$X$ is minimal Hausdorff iff $X$ is $H$-closed and semiregular.

and the fact that, by definition, $(X,\tau_2)$ is not minimal Hausdorff. 
For useful implications the inclusion goes the wrong way: $(X,\tau_1)$ $H$-closed means that every $\tau_1$-open cover has a finite subset whose closures cover $X$. But this says not too much about $\tau_2$-open covers. 
A possible example of this situation actually occurring is $X=[0,1]$ in $\tau_1$ the usual topology and $\tau_2$ the topology generated by $\{\mathbb{Q}\} \cup \tau_1$ which is a well-known example of a non-compact $H$-closed space. But there might be a counterexample too. 
A: Unfortunately, finer topologies won't necessarily be H-closed. For example, suppose $\langle X,\tau_1\rangle$ is a topological space that is H-closed, such that $X$ is infinite, and consider the discrete topology $\tau_2$ on $X.$ Since regular Hausdorff H-closed spaces are compact, then $\langle X,\tau_2\rangle$ cannot be H-closed, as the open cover singletons has no finite subcover. However, we trivially have $\tau_1\subset\tau_2.$
