$\DeclareMathOperator\cp{cp}$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 $\cp(\tau)$ as the one generated by
$$\{ K^c, \text{$K$ compact in $\tau$} \}.$$
Remark 0:derive some additional necessary conditions from the latter set forms a basis for $\cp(\tau)$. Indeed, a finite union of compacts is compact, which yields that this set is closed under finite intersection.following
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.
Observation: Let $\tau$ be a topology on $X$ and $\tau'$ a topology refining $\tau$. Suppose that $(X,\tau')$ is compact. Then any $\tau'$-closed set is $\tau$-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.
FromConsequences: Let $\tau^{\zeta}\subset \cp(\tau)$, we get that the latter must$(X,\tau)$ be Hausdorff. Unwinding the definitions, this amounts to saying that $\tau$ is compactly separated. Indeed, you can always suppose that the opens you use for Hausdorffness are in a certain basis for the topologytopological space admitting a $\beta$-structure $\tau^\xi$. Then:
$(X,\tau)$ is compact (as noted in the question).
$(X,\tau)$ is locally compact (in the sense that for every $x \in X$ there is a local base of compact neighborhoods). This follows from condition (3) on a $\beta$-space and the Observation.
$(X,\tau)$ is "c-separated": For every disjoint $C,D \subseteq X$ which are either closed or singletons, there exist compact $K,L \subseteq X$ such that $C \cap K = \emptyset$, $D \cap L = \emptyset$, and $K \cup L = X$. This follows from the fact that $(X,\tau^\xi)$ is Hausdorff, regular, and normal and the Observation.
$(X,\tau)$ is "c-completely separated": Let $C,D \subseteq X$ be disjoint and either closed or singletons. Then there exists a (not necessarily continuous) function $f: X \to [0,1]$ such that $f^{-1}(0) = C$, $f^{-1}(1) = D$, and $f^{-1}([a,b])$ is compact for every $a \leq b$. This follows from the fact that $(X,\tau^\xi)$ has the corresponding separation property and the Observation.
Now we show that they areNote also sufficient. We claim that $\cp(\tau)$ makesif the work. As we noticed, it is Hausdorff because $\tau$ is compactly separated. It is alsocollection of sets with compact: if complement forms a family $K_i^c $ covers $X$, the intersection of $K_i$ is emptytopology, and by a famous lemma a finite intersection of themthis this topology is empty. Note that we can test compactness using just opens from a basis, by the Alexander Theorem.
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 compactunique $K$ such that$\beta$-structure on $U \subset K \subset V$$(X,\tau)$. But then $K$this is closed in $\cp(\tau)$, which gives thatnot necessarily 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 properlycase. Feel free to edit :) hope it's correct!
