The standard entourages on minimal Cauchy filters form a separated uniformity

Because a uniformity is a proper filter on $X\times X$, its carrier $X$ is nonempty. Choose $x\in X$; then [L1] gives a minimal Cauchy filter associated to the principal filter at $x$, so $\widehat{X}$ is nonempty. For symmetric $E$, put $(\mathcal{F},\mathcal{G})\in \widehat{E}$ when some $A\in \mathcal{F}$ and $B\in \mathcal{G}$ satisfy $A\times B\subseteq E$.

Every $\mathcal{F}$ is $\widehat{E}$-close to itself: choose an $E$-small member of the Cauchy filter and use it on both sides. Thus each $\widehat{E}$ contains the nonempty diagonal of $\widehat{X}$. The relation $\widehat{E}$ is symmetric. If $\widehat{E}$ and $\widehat{D}$ have respective witnesses $A\times B$ and $C\times K$, then $(A\cap C)\times (B\cap K)\subseteq E\cap D$, so finite intersections are refined by the corresponding hatted intersection.

Choose a symmetric entourage $D$ with $D^{\circ 2}\subseteq E$. If $\mathcal{F}\widehat{D}\mathcal{G}$ via $A\times B\subseteq D$ and $\mathcal{G}\widehat{D}\mathcal{H}$ via $C\times K\subseteq D$, choose $b\in B\cap C$. Then $aDbDk$ for every $a\in A,k\in K$, so $A\times K\subseteq D^{\circ 2}\subseteq E$. Hence $\widehat{D}\circ \widehat{D}\subseteq \widehat{E}$.

Steps 2.1 and 2.2 show that the upward closure of the relations $\widehat{E}$ is a uniformity. To prove separation, suppose $\mathcal{F}\widehat{E}\mathcal{G}$ for every entourage $E$. Given $A\in \mathcal{F}$, minimality gives $\mathcal{F}=m(\mathcal{F})$ by [L1], so some $D[C]\subseteq A$ with $C\in \mathcal{F}$ and symmetric $D$. Choose an entourage $E\subseteq D$ and witnesses $P\in \mathcal{F},Q\in \mathcal{G}$ with $P\times Q\subseteq E$. Pick $c\in C\cap P$. Then $Q\subseteq E[c]\subseteq D[C]\subseteq A$, so $A\in \mathcal{G}$. Thus $\mathcal{F}\subseteq \mathcal{G}$; symmetry gives equality.

Therefore the standard relations form the asserted separated uniformity.