Let $X$ be a uniform space, $S\subseteq X$ and $f:S\to \mathbb{R}$ bounded uniformly continuous, then there exists a uniformly continuous extension of $f$ to $X$. (Katětov, 1951)
I am looking for proof of above theorem with uniform spaces defined using pseudometrics, or just a modern proof of it. Katětov's argument with entourages is not very digestible to me.
$\DeclareMathOperator{\R}{\mathbb R} \DeclareMathOperator{\eps}{\varepsilon}$ If you prefer to define uniformities in terms of a family $D$ of pseudometrics you can reduce the theorem to pseudometric spaces $(X,d)$.
Indeed, for every $n\in\mathbb N$ there are $d_n\in D$ and $\delta_n>0$ with $$ d_n(x,y) <\delta_n \, \Rightarrow \,|f(x)-f(y)|< 1/n $$ for all $x,y\in S$. Passing to finite maxima we may assume $d_n\le d_{n+1}$. Then $$ d(x,y)=\sup\{d_n(x,y) \wedge 1/n: n\in\mathbb N\} $$ (where $a\wedge b=\min\{a,b\}$) is a pseudometric compatible with the uniformity such that $f$ is uniformly continuous on $S$ with respect to $d$.
The pseudometric case was proved by Whitney as well as MacShane along the following lines (the following is from a manuscript of mine where the function is $\varphi:A\to\mathbb R$).
We define the minimal modulus of continuity of $\varphi$ for $t\in (0,\infty)$ as $$ \omega_0(t)=\sup\{|\varphi(x)-\varphi(y)|: x,y\in A \text{ with } d(x,y)\le t\}. $$ This function $\omega_0$ is increasing and bounded by $2c$ if $c\ge 0$ is a bound for $\varphi$, and the uniform continuity of $\varphi$ implies $\omega_0(t)\to 0$ for $t\to 0$. The concave majorant of $\omega_0$ is $$ \omega(t)=\inf\{at+b: a,b\ge 0 \text{ with } \omega_0(s)\le as+b\text{ for all }s>0\} $$ (instead of the boundedness of $\omega_0$ it would be enough to assume a sub-linear growth). For $t,q>0$, $\lambda\in (0,1)$, the convex combination $r=\lambda t+(1-\lambda)q$, and $\eps>0$, $\omega(r)+\eps$ isn't a lower bound for the set corresponding to $r$ so that there are $a,b\ge 0$ with $as+b\ge \omega_0(s)$ on $(0,\infty)$ and $ar+b<\omega(r)+\eps$. This implies $$ \lambda\omega(t)+(1-\lambda)\omega(q)\le \lambda(at+b)+(1-\lambda)(aq+b)=ar+b\le \omega(r)+\eps $$ so that $\eps\to 0$ yields that $\omega$ is indeed concave on $(0,\infty)$. Let us next show that $\omega(t)\to 0$ for $t\to 0$: For $\eps>0$ we take $\delta>$ with $\omega_0(t)\le \eps$ for all $t\le\delta$. For $a=2c/\delta$ we have $\omega_0(s)\le as+\eps$ for $s\in (0,\infty)$ and hence $\omega(t)\le at+\eps\le 2\eps$ for all $t$ small enough. We extend $\omega$ to $[0,\infty)$ by $\omega(0)=0$. This concave and increasing function is then subadditive: For $t,u\ge 0$ not both $0$ we set $\lambda=t/(t+u)$ and add the inequalities \begin{align*} \omega(t) & =\omega(\lambda(t+u)+(1-\lambda)0)\ge \lambda\omega(t+u) \text{ and } \\ \omega(u) &=\omega((1-\lambda)(t+u)+\lambda 0)\ge (1-\lambda)\omega(t+u). \end{align*} Since $\omega$ is positive, increasing and sub-additive with $\omega(0)=0$ we obtain a new semi-metric $\tilde d(x,y)=\omega(d(x,y))$ such that $\varphi:A\to \R$ is a weak $\tilde d$-contraction (i.e., it has Lipschitz constant $1$). The theorem of McShane-Whitney (i.e., the simple formula $f(x)=\inf\{\varphi(a)+\tilde d(x,a): a\in A\}$) thus yields an extension $f:X\to \R$ which is a weak contraction with respect to $\tilde d$ and hence uniformly continuous with respect to $d$.
