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 $E_{\mathcal{W}}[x]=\mathrm{St}(x,\mathcal{W})$ 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 $E_{\mathcal{W}}[x]=\mathrm{St}(x,\mathcal{W})\subseteq \mathrm{St}(W_0,\mathcal{W})\subseteq O$.

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