Let $X$ be a locally compact Hausdorff space and suppose that $C_0(X)$ is separable w.r.t the uniform metric. Then $X$ is metrizable and separable.
I'm reading a proof of this statement that goes like this: let $X^*$ denote the one-point compactification of $X$ and note that $C(X^*)$ is separable w.r.t the uniform metric. By some standard results, $X^*$ is thus metrizable, hence $X$ is metrizable and separable.
I don't get why "$C_0(X)$ is separable" implies "$C(X^*)$ is separable". Letting $(f_n)$ be a dense sequence in $C_0(X)$, each $f_n$ can be continuously extended to a function $f_n^*\in C(X^*)$ with $f_n^*(\infty)=0$. Given $g\in C(X^*)$ and $\epsilon>0$, if $g(\infty)=0$ then $g$ is clearly approximated by some $f_n^*$. However, if $g(\infty)\neq 0$, I'm stuck.
Note that $C(X^*)=C_0(X)\oplus\Bbb R$ and the product of separable spaces is separable. More concrete: let $g\in C(X^*)$. Let $\alpha\in\Bbb Q$ such that $|g(\infty)-\alpha|<\frac{\varepsilon}{2}$ and $n$ such that $\|(g-g(\infty))-f_n\|<\frac{\varepsilon}{2}$ (note that $g-g(\infty)\in C_0(X)$). Then $$\|g-(f_n+\alpha)\|\leq\|(g-g(\infty))-f_n\|+|g(\infty)-\alpha|<2\frac{\varepsilon}{2}=\varepsilon$$ Thus the countable set $(f_n+\alpha)_{n\in\Bbb N,\alpha\in\Bbb Q}$ is dense in $C(X^*)$.