It helps to have the concept of a "little Lipschitz function". A Lipschitz function $f: X \to \mathbb{R}$ is *little Lipschitz* if $\frac{|f(p) - f(q)|}{\rho(p,q)} \to 0$ as $p,q \to 0$. Notation: ${\rm Lip}(X)$ is the set of real-valued Lipschitz functions on $X$ and ${\rm lip}(X)$ is the set of real-valued little Lipschitz functions on $X$. The usual norm is $\|f\|_L = \max(L(f), \|f\|_\infty)$ where $L(f)$ is the Lipschitz number of $f$.

Also, for any metric space $X$ and $0 < \alpha < 1$ let $X^\alpha$ be the same set equipped with the metric $\rho^\alpha(p,q) = (\rho(p,q))^\alpha$. Then $C^{0,a}(\overline{\Omega}) = {\rm Lip}(\overline{\Omega}^\alpha)$.

The facts you want here are that (1) ${\rm Lip}(X) \subset {\rm lip}(X^\alpha)$ for any compact $X$ and any $0 < \alpha < 1$ and (2) $C^\infty(\overline{\Omega})$ is dense in ${\rm lip}(\overline{\Omega}^\alpha)$. Since $X^\beta = (X^\alpha)^{\beta/\alpha}$, this yields $C^\infty(\overline{\Omega}) \subset {\rm Lip}(\overline{\Omega}^\alpha) \subset {\rm lip}(\overline{\Omega}^\beta)$, with the first space being dense in the last, so the first is dense in the second which is what you asked about.

For those two facts see Example 3.1.7 and Theorem 4.4.2 of my book *Lipschitz Algebras*. Density follows from Theorem 4.4.2 because $C^\infty(\overline{\Omega})$ separates points "uniformly"; this can be seen by checking that for any $x,y \in \overline{\Omega}$ the function $z \mapsto a^{\alpha-1}\cdot\max(a - |x - z|, 0)$ with $a = |x - y|$ has Lipschitz number $1$ in ${\rm Lip}(\overline{\Omega}^\alpha)$ and separates $x$ and $y$ to their full distance; smooth this to get a $C^\infty$ function which has Lipschitz number $\leq 1$ and separates $x$ and $y$ to nearly their full distance.