Isolated Points of Countable H-Closed Spaces 
Let $X$ be a countable H-closed space, and $D:=$ Set of isolated points of $X$. Then, $D$ is dense in $X$

I attempted to do this by contradiction $-$ if $D$ is not dense in $X$, then $\exists x\in X-\bar D$ and an open neighbourhood $U$ of $x$ such that $U\subseteq X-\bar D$. From here, I have to get an open cover for which no finite subcollection has closures covering $X$. But, I wasn't able to construct it. Any help is appreciated!
 A: $\newcommand{\cl}{\operatorname{cl}}$First show that any countable $H$-closed space $X$ has an isolated point. Suppose that $X$ has no isolated points; we will get a contradiction by producing an open filter base with no cluster point. ($X$ is $H$-closed iff every open filter on $X$ has a cluster point.) The construction is recursive.
Let $X=\{x_n:n\in\Bbb N\}$. Let $y_0\in X\setminus\{x_0\}$; there are disjoint open sets $U_0$ and $V_0$ such that $x_0\in U_0$ and $y_0\in V_0$. $V_0$ is infinite, since $X$ has no isolated points, so there is a point $y_1\in V_0\setminus\{x_1\}$, and there are disjoint open sets $U_1$ and $V_1$ such that $x_1\in U_1$ and $y_1\in V_1\subseteq V_0$. Given $x_k,y_k,U_k$, and $V_k$ for $k\le n$, choose $y_{n+1}\in V_n\setminus\{x_{n+1}\}$, and let $U_{n+1}$ and $V_{n+1}$ be disjoint open sets such that $x_{n+1}\in U_{n+1}$, and $y_{n+1}\in V_{n+1}\subseteq V_n$. Now verify that the filter generated by the filter base $\{V_n:n\in\Bbb N\}$ has no cluster point in $X$; this is where you’ll use the sets $U_n$.
Now let $D$ be the set of isolated points of $X$, and if $\cl D\ne X$, let $G=X\setminus\cl D$. $G$ has no isolated points, so the previous construction can be carried out entirely in $G$ if we choose $y_0\in G\setminus\{x_0\}$ and $V_0\subseteq G$ at the start, and we get the same contradiction.
