 Solution

The real Stone-Weierstrass theorem says that a real subalgebra $A\subset C(K,\mathbb{R})$ which contains the constants and separates points is dense in $C(K,\mathbb{R})$ for the uniform norm, when $K$ is compact Hausdorff space. If $A$ is also closed, $A=C(K,\mathbb{R})$. The complex version needs closure under complex conjugation; that extra condition is unnecessary here.