I am looking for a reference to a proof for the following result: let $m \in \mathbb{N}, \alpha \in (0,1)$ and $\Omega \subset \mathbb{R}^m$ be a bounded open set. Let $f_n: \Omega \to \mathbb{R}$ be a sequence of $C^{0, \alpha}$ functions that converges uniformly to $f : \Omega \to \mathbb{R}$. Then $f \in C^{0, \alpha}(\Omega)$ and $f_n \to f$ as $n \to \infty$ in $C^{0, \alpha}$ topology.
This result does not have a complicated/hard proof. It is immediate to see that $f \in C^{0, \alpha}(\Omega),$ and the inequality $$\begin{align*} \frac{|(f(x) - f_n(x)) - (f(y) - f_n(y)) |}{||x-y||^\alpha} &= \lim_{m \to \infty} \left( \frac{|(f_m(x) - f_n(x)) - (f_m(y) - f_n(y)) |}{||x-y||^\alpha} \right) \\ &\leq \limsup_{m \to \infty}(||f_m - f_n||_{\alpha}) \to 0 \text{ as } n \to \infty, \end{align*}, $$ implies that $f_n \to f$ in $C^{0, \alpha}$ topology, where $||\cdot||_\alpha$ is the $\alpha-$Holder norm. However, I did not manage to find a book where this is proven explicitly.
I don't understand your proof, why is the $\alpha$ norm going to $0$? Uniform convergence just tells you that $\|f_n-f\|_{L^\infty}$ converges to $0$.
And actually what you are asking is false. Just take $f(x) = |x|^c$ with $0<c<\alpha$. Then $f\notin C^{0,\alpha}$. But if you take $f_n(x) = \max(1/n,f(x))$ for example, then $f_n\in C^{0,1}\subset C^{0,\alpha}$ and $f_n$ converges uniformly to $f$, since $\|f_n-f\|_{L^\infty}<1/n$.