This is an exercise from Herrlich's Axiom of choice, and I'm a bit struggling with it.
I want to show that a topological space $X$ is compact Hausdorff if and only if it is H-closed and regular.
NB : I don't know what the convention is, but here H-closed means that for any Hausdorff $Y$ such that there is an (continuous) embedding $X\to Y$, then the image of $X$ is closed.
I managed to show the "compact$\implies$ H-closed" part, through the filter-compact characterization of compact spaces; but the other part I couldn't do.
I can't see how to use H-closed ness to prove that $X$ has the Borel-Lebesgue property, and so I think that using the filter characterization would be the best; but given a filter $\mathcal{F}$ with no cluster point, I don't see what space I could embed $X$ in that would give me nice things. Any indications ?
Use that $X$ is $H$-closed iff every open cover has a finite subfamily whose union is dense, then use regularity to go to compactness.
See my note here for a proof of this.
Another way is to use the filter characterisation of $H$-closedness:
$X$ is $H$-closed iff every open filter $\mathscr{F}$ of $X$ (i.e. a filter in the poset of open sets of $X$) has an adherence point. (i.e. $\bigcap \{\overline{F}: F \in \mathscr{F} \} \neq \emptyset$).
Then compact implies $H$-closed is then also clear (And compact plus Hausdorff implies regular) (as the closures form a FIP family). And if $X$ is $H$-closed and regular, take any open cover $\mathscr{U}$ of $X$. If it has no finite subcover, refine it (using regularity) by a cover of regular closed sets having no finite subcover as well, and then consider the open filter generated by the complements of finite unions of subfamilies of that family...