Every uniform space has a Hausdorff completion with dense canonical image, and the canonical map is a uniform embedding exactly when the original uniformity is separated

Take $\widehat X$ to be the uniform space of minimal Cauchy filters and take $\eta$ from [L2].

It is complete and separated by [L1], and $\eta$ is uniformly continuous with dense image by [L2]. It remains to verify that the pullback uniformity is not strictly coarser than the original one. Given an entourage $E$ of $X$, choose a symmetric $D\subseteq E$. If $(\eta(x),\eta(y))\in\widehat D$, witnesses $A\in\eta(x)$ and $B\in\eta(y)$ satisfy $A\times B\subseteq D$. Every member of the minimal point filter $\eta(x)$ contains $x$, and every member of $\eta(y)$ contains $y$; therefore $(x,y)\in D\subseteq E$. Thus $(\eta\times\eta)^{-1}[\widehat D]\subseteq E$. Together with uniform continuity, this is exactly the pullback condition in [L3], so $\eta$ is a Hausdorff completion.

If $\eta(x)=\eta(y)$, step 2.1 puts $(x,y)$ in every entourage of $X$. Conversely, if $(x,y)$ belongs to every entourage of $X$, uniform continuity puts $(\eta(x),\eta(y))$ in every entourage of $\widehat X$; separatedness of $\widehat X$ gives $\eta(x)=\eta(y)$.

Step 3.1 says that $\eta$ is injective exactly when $\mathcal U$ is separated. When injective, the two directions of the pullback condition in step 2.1 say precisely that the corestriction $X\to\eta[X]$ and its inverse are uniformly continuous, so $\eta$ is a uniform embedding. Conversely every uniform embedding is injective.

This proves the completion assertion and the exact embedding criterion.