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 $D^{\circ2}\subseteq E$. 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 $A\times K\subseteq D^{\circ2}\subseteq E$. 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.