In the paper A survey of D-spaces by Gary Gruenhage it is written that it is easily seen that any countably compact D-space is compact. However I'm not able to show it. Here is my attempt to prove this claim: Let $(X, \tau)$ be a countably compact D-space with topology $\tau$. Consider a function $N: X \rightarrow \tau$ with $x \in N(x)$ for every $x \in X$ (neighborhood assignment function). I think that to prove compactness of $X$ it is enough to show that open cover $\{N(x):x \in X\}$ of $X$ has finite subcover. Because $X$ is a D-space, there is a closed discrete subset $D$ of $X$ such that $\{N(x):x \in D\}$ covers $X$. It seem to me that to complete the proof (using countable compactness of $X$) I need to use the fact that $D$ is countable. However it is not true in general that close discrete set is countable. Thank you for your help! Michal Čihák
Note 1. The considered spaces are regular and $T_1$.
Note 2. A space $X$ is a D-space if whenever one is given a neighborhood $N(x)$ of $x$ for each $x \in X$, then there is a closed discrete subset $D$ of $X$ such that $\{N(x):x \in D\}$ covers X.