How do I prove the converse of Stone-Weierstrass theorem? Let $X$ be a locally compact Hausdorff space. Let $\bar \rho$ be the uniform metric on $\mathbb{R}^X$ and $\mathscr{A}$ be an $\mathbb{R}$-subalgebra of $C_0(X,\mathbb{R})$ which is dense in $(C_0(X,\mathbb{R}),\bar \rho)$.
In this case, how do I prove that $\mathscr{A}$ separates points on $X$ and $\mathscr{A}$ vanishes nowhere?
I can prove this by applying Urysohnn's Lemma if $X$ is compact, but cannot handle this when $X$ is locally compact.
 A: It suffices to prove that $C_0(X,\mathbb{R})$ separates points and vanishes nowhere.
Let $\bar X$ be the one-point compactification of $X$ so that $\bar X$ is compact Hausdorff, hence $T_4$, in which Urysohnn's lemma is applicable.
Now, define $\mathscr{C}\triangleq \{f\in C(\bar X,\mathbb{R}): f(\infty)=0\}$.
Note that $\mathscr{C}$ is an $\mathbb{R}$-subalgebra of $C(\bar X,\mathbb{R})$ and there is an $\mathbb{R}$-algebra isomorphism $\phi:\mathscr{C}\rightarrow C_0(X,\mathbb{R}):f\mapsto f\upharpoonright X$.

Claim1. $C_0(X,\mathbb{R})$ vanishes nowhere.

Proof> 
Fix $x\in X$.
Then, there exists $g\in C(Y,[0,1])$ such that $g(x)=1$ and $g(\infty)=0$ by Urysohnn's Lemma.
Hence $g\in \mathscr{C}$ and $g(x)=1$.
Thus $\phi^{-1}(g)\in C_0(X,\mathbb{R})$ and $\phi^{-1}(g)(x)\neq 0$. Q.E.D.

Claim2. $C_0(X,\mathbb{R})$ separates points.

Proof>
Let $x,y$ be distinct points in $X$.
Then, there exists $g\in C(\bar X,[0,1])$ such that $g(x)=1$ and $g(y)=0$, hence $g(x)\neq g(y)$.
Moreover, there exist $h,k\in \mathscr{C}$ such that $h(x)\neq 0$ and $k(y)\neq 0$.
Define $u\triangleq gk-g(x)k$ and $v\triangleq gh-g(y)h$.
Then, it can be directly checked that $u,v\in \mathscr{C}$ and $u(y)\neq 0$ and $v(x)\neq 0$.
Now, set $f\triangleq \frac{u}{u(y)} + 2\frac{v}{v(x)}$.
Then, $f\in \mathscr{C}$ and $f(x)\neq f(y)$.
Hence the inverse of $f$ under $\phi$ separates $x,y$. Q.E.D.
