I'm not sure why the space $(Y,\mathcal{V})$ described below is completely regular.
I start by taking $(S,\mathcal{U})$ and defining an equivalence relation $s\sim t\iff f(s)=f(t)$ for every continuous $f\colon S\to\mathbb{R}$. Let $Y$ be the set of such equivalence classes, and $\pi$ be the function mapping $s\in S$ to its equivalence class. So for continuous $f\colon S\to\mathbb{R}$, there is a unique $\phi(f)\colon Y\to\mathbb{R}$ such that $\phi(f)(\pi(s))=f(s)$. Then equip $Y$ with the weakest topology $\mathcal{V}$ such that each $\phi(f)$ is continuous, so every closed set in $Y$ has form $\bigcap_{i\in I}\phi(f_i)^{-1}(F_i)$ for some family $\{F_i\}$ of closed subsets of $\mathbb{R}$ and $\{f_i\}$ continuous.
With Brian M. Scott's aid, I've seen $(Y,\mathcal{V})$ is Hausdorff. But why is $(Y,\mathcal{V})$ completely regular? I take any closed set $A$ in $Y$, so $ A=\bigcap_{i\in I}\phi(f_i)^{-1}(F_i) $ for some index set $I$. I pick some point $b\not\in A$, so there is some $i\in I$ such that $b\not\in \phi(f_i)^{-1}(F_i)$, that is, $\phi(f_i)(b)\not\in F_i$. Based on that, why is there some continuous function separating $A$ and $b$, with $A$ mapping into $\{0\}$ and $b$ mapping to $1$?
Is it then an easy consequence that the map $\pi$ is continuous as well? If not, perhaps I'll post that as a separate question later. Thanks!