The Samuel uniformity is totally bounded

For each $f\in F$, [L2] supplies a finite set $A_f\subseteq[0,1]$ such that every value of $f$ is within $\varepsilon/3$ of some member of $A_f$.

The product $A:=\prod_{f\in F}A_f$ is finite, and for $a\in A$ let $C_a$ be the set of $x\in X$ with $|f(x)-a_f|<\varepsilon/3$ for every $f\in F$.

The index set $A':=\{a\in A:C_a\ne\varnothing\}$ of nonempty cells is a finite subset of $A$. Choose a natural $n$ and a bijection $e:n\to A'$, form the explicitly $n$-indexed family $i\mapsto C_{e(i)}$, and use [L3] to choose $c_{e(i)}\in C_{e(i)}$; let $C$ be the set of chosen points.

If $x\in X$, choose $a\in A$ with $x\in C_a$ using step 1.1; then $a\in A'$ and $|f(x)-f(c_a)|<2\varepsilon/3<\varepsilon$ for every $f\in F$, so $x\in E(F,\varepsilon)[c_a]$.

Thus $C$ is a finite net for each basic Samuel entourage. Every Samuel entourage contains one of these basic entourages, so the same finite centres cover it; when $F=\varnothing$, use the empty centre set if $X=\varnothing$ and any singleton centre otherwise. Hence $\mathcal U_S$ is totally bounded.