The exercise: Let $\Omega$ be a compact Hausdorff metrizable space. Show that the subspace $C^{\alpha}(\Omega, \mathbb{R})$ of $\alpha$-Hölder continuous function is dense on $C(\Omega, \mathbb{R})$.
Now, a theorem I have at my disposal is:
(Stone-Weierstrass Theorem) Let $X$ be a compact Hausdorff space. Let $S \subset C(X, \mathbb{R})$ obey:
i) S is a subalgebra of $C(X, \mathbb{R})$, that is, $f, g \in S$ and $\lambda \in \mathbb{R} \implies f + g, fg, \lambda f \in S$
ii) $S$ separates points , i.e, for each $x, y \in \Omega$ such that $x \neq y$, there exists $f \in C(\Omega, \mathbb{R})$ such that $f(x) \neq f(y)$
iii) $\mathbb{1}$, the constant function $\equiv 1$ belongs to $S$
then $\overline { S } ^ { \| \cdot \| _ { 0 } } = C ( \Omega , \mathbb { R } )$, i.e, $S$ is dense in $C(\Omega, \mathbb{R})$
My effort:
i) The only non-trivial part is showing that if $f, g$ are $\alpha$-Holder continuous then $fg$ as well. Indeed, we have:
\begin{align} \frac{|f(x)g(x) - f(y)g(y)|}{d(x,y)^{\alpha}} &\leq \frac{|f(x)| |g(x) - g(y)|}{d(x,y)^{\alpha}} + \frac{|g(y)||f(x) - f(y)|}{d(x,y)^{\alpha}} \\ &\leq \|f \| \text{Hol}_{\alpha}(g) + \|g \| \text{Hol}_{\alpha}(f)\end{align}
and then we're done because this implies $\text{Hol}_{\alpha}(fg)$ is finite, as desired.
ii) Let $x, y \in \Omega$ be given such that $x \neq y$. Then if we define $f_{x}(y) = d(x, y)^{\alpha}$ we have $0 = f_{x}(x) \neq f_{x}(y)$ because $x \neq y$. All that's left to show is that $\text{Hol}_{\alpha}(f_x)$ is finite. This is indeed the case because $\text{Hol}_{\alpha}(f_x) = 1$.
iii) This one is trivial because $\text{Hol}_{\alpha}(1) = 0$.
Is all of this correct? Is there something I could've done better? In case there's anything fundamentally wrong, a hint would really help!