Regular closed subset of H-closed space An H-closed space $X$ is a topological space which is closed in any Hausdorff space in which it is embedded. A well-known characterization is that $X$ is H-closed iff every open cover of $X$  has a finite proximate subcover, i.e. a finite subcollection whose union is dense. 
I need to show that this property is hereditary to regular closed subsets. I tried to do something analogous to the proof of closed subset of Compact spaces are compact. But got stuck. Any help is appreciated.
 A: Let $X$ be H-closed, and let $F$ be a regular closed set in $X$. Let $\mathscr{U}$ be a relatively open cover of $F$. For each $U\in\mathscr{U}$ there is an open $V_U$ in $X$ such that $U=F\cap V_U$; let 
$$\mathscr{V}=\{X\setminus F\}\cup\{V_U:U\in\mathscr{U}\}\;.$$
$\mathscr{V}$ is an open cover of $X$, so it has a finite proximate subcover $\mathscr{V}_0$. Let 
$$\mathscr{U}_0=\{U\in\mathscr{U}:V_U\in\mathscr{V}_0\}\;;$$
clearly $\mathscr{U}_0$ is a finite subset of $\mathscr{U}$. Since $\operatorname{cl}(X\setminus F)\cap\operatorname{int}F=\varnothing$, and $\bigcup\mathscr{V}_0$ is dense in $X$, $\bigcup\{V_U:U\in\mathscr{U}_0\}$ must be dense in $\operatorname{int}F$, and hence $\bigcup\mathscr{U}_0$ must be dense in $\operatorname{int}F$. Thus,
$$F=\operatorname{cl}\operatorname{int}F\subseteq\bigcup_{U\in\mathscr{U}_0}\operatorname{cl}U\subseteq F\;,$$
$\mathscr{U}_0$ is a proximate subcover of $\mathscr{U}$, and $F$ is H-closed.
It is also true that a space $X$ is H-closed iff every open filter in $X$ has a cluster point, and we can use this characterization instead. Let $\mathscr{U}$ be a relatively open filter on $F$, and let $\mathscr{B}=\{U\cap\operatorname{int}F:U\in\mathscr{U}\}$. Clearly $U\cap\operatorname{int}F\ne\varnothing$ for each $U\in\mathscr{U}$, so $\mathscr{B}$ is an open filterbase in $X$. $X$ is H-closed, so the filter $\mathscr{V}$ generated by $\mathscr{B}$ has a cluster point $x\in X$, which is evidently also a cluster point of $\mathscr{U}$. And $\operatorname{int}F\in\mathscr{V}$, so every nbhd of $x$ meets $\operatorname{int}F$, and therefore $x\in\operatorname{cl}\operatorname{int}F=F$, so that $\mathscr{U}$ has a cluster point in $F$.
