Let $X$ be a topological space with a dense subset $D\subseteq X$. Suppose that every open cover of $X$ has a finite subfamily which covers $D$. Can I conclude that $X$ itself is compact?

The answer is clearly negative in general: one can take any space $X_0$ and form a new space $X:=X_0\cup\{\infty\}$ by adjoining a new point $\infty$ such that the nonempty open sets are those of the form $\{\infty\}\cup U$ for $U$ open in $X_0$. Then the closure of $\{\infty\}$ is the whole space, so that the condition trivially holds with $D=\{\infty\}$. Now take any non-compact $X_0$.

Nevertheless, compactness of $X$ indeed follows if $X$ is completely regular. One way to show this is to argue that the condition is equivalent to every net in $D$ having a cluster point, and then using uniformizability to approximate an arbitrary net in $X$ by one in $D$. But of course, a finite subfamily which covers $D$ does not necessarily cover $X$ yet, even though a finite subcover is guaranteed to exist.

So, perhaps surprisingly, the answer is positive in particular for any metric space $X$.

Now my question is:

Are there weaker separation conditions than complete regularity on $X$ which guarantee that the finite subcover property for $X$ only needs to be checked on a dense subset $D\subseteq X$?