The covers admitting an open refinement form a compatible uniform-cover structure on a nonempty compact Hausdorff space; in particular every open cover is uniform

Let $\mathfrak C$ be the covers admitting an open refinement. It is nonempty, since $\{X\}$ is open.

Coarsening preserves membership in $\mathfrak C$, and two open refinements have their intersection cover as a common open refinement.

For $\mathcal V\in\mathfrak C$, take an open refinement and then its finite open star-refinement from [L1]; this is a star-refinement still witnessing membership in $\mathfrak C$.

Thus $\mathfrak C$ satisfies [L2]. Since every open cover refines itself, every open cover belongs to $\mathfrak C$.

The entourage uniformity recovered from $\mathfrak C$ by [L3] induces the original topology. If $\mathcal V\in\mathfrak C$, choose an open refinement $\mathcal W$; then $E_{\mathcal W}[x]=\operatorname{St}(x,\mathcal W)$ contains an open member through $x$, so the recovered entourage balls are neighbourhoods in the original topology. Conversely, if $x\in O$ with $O$ open, the open cover $\{O,X\setminus\{x\}\}$ belongs to $\mathfrak C$, since Hausdorffness makes $\{x\}$ closed. By [L1] choose a finite open star-refinement $\mathcal W$, which belongs to $\mathfrak C$, and choose $W_0\in\mathcal W$ containing $x$. The star of $W_0$ lies in $O$, rather than in $X\setminus\{x\}$, and therefore $E_{\mathcal W}[x]=\operatorname{St}(x,\mathcal W)\subseteq\operatorname{St}(W_0,\mathcal W)\subseteq O$.

Hence the structure is compatible with the given topology, and every open cover is uniform.