= Solution
The real <Stone-Weierstrass theorem> says that if $X$ is compact Hausdorff and $A\subset C(X,\mathbb R)$ is a subalgebra containing the constants and separating points, then $A$ is uniformly dense in $C(X,\mathbb R)$.
Let $\overline A$ be the uniform closure. <Polynomial> approximation to $\sqrt t$ on a bounded interval shows that $|f|\in\overline A$ whenever $f\in\overline A$. Hence $\overline A$ is closed under
$$
\max(f,g)=\frac{f+g+|f-g|}{2},\qquad
\min(f,g)=\frac{f+g-|f-g|}{2}.
$$
Given $h\in C(X)$, separation and the constants provide, for each $x,y$, a <function> in $A$ agreeing with $h$ at $x$ and $y$. Compactness first combines finitely many such <functions> by minima to obtain one that agrees at $x$ and lies below $h+\varepsilon$ everywhere; a second finite cover and maxima produces $f\in\overline A$ with $|f-h|<\varepsilon$. Thus $h\in\overline A$.
The space $C_b(\mathbb R)$ is Banach because a uniform Cauchy <sequence> converges uniformly to a bounded <continuous function>. Compactness is essential: the algebra of bounded <continuous functions> having finite <limits> at both $+\infty$ and $-\infty$ contains constants and separates points, for example using $\tanh x$, but its uniform closure has the same limiting property. It cannot uniformly approximate $\sin x$, so it is not dense in $C_b(\mathbb R)$.
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