A nonempty compact Hausdorff space carries exactly one compatible uniformity

Apply [L2] to the cover structure of [L1] to obtain one compatible entourage uniformity.

Let $\mathcal U$ be any compatible uniformity. Each entourage-ball cover admits an open refinement because every ball is a neighbourhood in the induced topology, so every cover uniform for $\mathcal U$ admits an open refinement.

Conversely, let $\mathcal O$ be an open cover and take a finite open star-refinement $\mathcal W$ by [L5]. Form the family of all open sets $N$ for which there are $x\in N$, $W\in\mathcal W$, and a symmetric entourage $D$ satisfying $N\subseteq D[x]$ and $D^{\circ2}[x]\subseteq W$. This family covers $X$: given $x$, first take $W\in\mathcal W$ containing it, then use compatibility and a symmetric square root to obtain such $D$ and an open neighbourhood $N\subseteq D[x]$. Compactness gives finitely many witnesses $(N_i,x_i,W_i,D_i)$ covering $X$. Put $D=\bigcap_iD_i$. If $y\in N_i$ and $z\in D[y]$, then symmetry gives $x_iD_i yD_i z$, so $z\in D_i^{\circ2}[x_i]\subseteq W_i$. Hence the $D$-ball cover refines $\mathcal W$, and therefore refines $\mathcal O$. Thus every open cover is uniform for $\mathcal U$.

By steps 1.2 and 1.3, the cover structure associated to $\mathcal U$ consists exactly of the covers admitting an open refinement, which is the structure in [L1].

The dictionary [L2] then recovers the same entourage uniformity from either structure, proving uniqueness.