Let $E$ be a complete separable metric space, $\mathcal C\subseteq C_b(E)$ be a point-separating$^1$ algebra$^2$, $\mu_i$ be a probability measure on $(E,\mathcal B(E))$ and $\varepsilon>0$.
We know that there is a compact $K\subseteq E$ with $\mu_i(K^c)<\varepsilon$. Let $f\in C_b(E)$. By the Stone-Weierstrass theorem, $\{\left.g\right|_K:g\in\mathcal C\}$ is dense in $C(K)$ and hence there is a $(f_n)_{n\in\mathbb N}\subseteq\mathcal C$ with $$\sup_{x\in K}\left|f_n(x)-f(x)\right|\xrightarrow{n\to\infty}0\tag1.$$
I would like to show$^3$ $$(\mu_1-\mu_2)f\le c\varepsilon\tag2$$ for some $c\ge0$.
I've found a proof which considers $(\mu_1-\mu_2)fe^{-\varepsilon f^2}$. They claim that "since $\mathcal C$ is an algebra, $f_ne^{-\varepsilon f_n^2}$ can be approximated by functions from $\mathcal C$ and so $(\mu_1-\mu_2)f_ne^{-\varepsilon f_n^2}\to 0$ as $n\to\infty$.
Why does this claim hold? Why does it follow that $f_ne^{-\varepsilon f_n^2}$ can be approximated by functions from $\mathcal C$?
$^1$ i.e. for all $x,y\in E$, there is a $f\in\mathcal C$ with $f(x)\ne f(y)$.
$^2$ i.e. $1\in\mathcal C$, $\alpha f\in\mathcal C$ for all $f\in\mathcal C$ and $g+h,gh\in\mathcal C$ for all $g,h\in\mathcal C$.
$^3$ For simplicity, write $$\nu h:=\int h\:{\rm d}\nu$$ whenever $\nu$ is a signed finite measure on $(E,\mathcal E)$ and $h\in\mathcal L^1(\nu)$.