We will show that $\tau$ admits such a refinement iff it is compact, locally compact and *compactly separated*, which means:

For every distinct $x,y$, there exist two compact sets $K_x, K_y$ such that
$$ x \not \in K_x, y \not \in K_y, K_x \cup K_y = X$$

Given a compact topology $\tau$ define the associate compact complement topology 

$$cp(\tau) :=\{ K^c, K \text{ compact in } \tau \}$$

Remark 1: $cp(\tau)$ refines $\tau$: if $U$ is an open in the latter, its complement is closed in a compact ($X$), thus it is compact. 

Remark 2: a closed set of a compact refinement $\tau'$ is compact in $\tau$. Indeed, it is compact in $\tau'$ because it is closed in a compact, and so it is compact also in $\tau$ because the identity $\tau' \to \tau$ is continuous.


Now let's show that such conditions are necessary. Suppose we have such a refinement $\tau^{\zeta}$.

Compactness has been shown. Analogously, local compactness follow from remark 2. 

From $\tau^{\zeta}\subset cp(\tau)$, we get that the latter must be Hausdorff. Unwinding the definitions, this amounts to say that $\tau$ is compactly separated.

Now we show that they are also sufficient. We claim that $cp(\tau)$ makes the work. As we noticed, it is Hausdorff because $\tau$ is compactly separated. It is also compact: if a family $K_i^c $ covers $X$, the intersection of $K_i$ is empty, and by a famous lemma a finite intersection of them is empty.

Let's get to "regularity": take $x \in V$, the latter being open. By local compactness, we know that there exist an open $U$ that contains $x$ and a compact $K$ such that $U \subset K \subset V$. But then $K$ is closed in $cp(\tau)$, which gives that the $cp(\tau)$ closure of $U$ is contained in $V$, and we are done.

Sorry if it is a mess, I didn't have time to write it properly. Feel free to edit :) hope it's correct!