This is to clarify that Lemma 24.3 is valid if $\Omega$ is Lipschitz. More generally, we have the following.
Lemma. Let $\Omega\in\mathbb{R}^n$ be a bounded domain with $C^{0,\alpha}$ boundary for some $1<\alpha\leq1$. Let $u\in C^1(\Omega)$ with all its partial derivatives bounded. Then $u\in C^{0,\alpha}(\Omega)$, i.e.,
$$
[u]_{\alpha}=\sup_{x,y\in\Omega}\frac{|u(x)-u(y)|}{|x-y|^\alpha}<\infty.
$$
Proof. Let $\{U_k\}$ be a (sufficiently fine) finite open cover of $\Omega$, such that in each $U_k$, the boundary of $\Omega$ is a rotated graph of a $C^{0,\alpha}$ function.
Then note that there exists $\varepsilon>0$ such that $\{|x-y|<\varepsilon\}\subset\bigcup_kU_k\times U_k$,
so that it suffices to prove
$$
\sup_{x,y\in U_k}\frac{|u(x)-u(y)|}{|x-y|^\alpha}<\infty,
$$
for each $k$. Pick $x,y\in U_k$, and suppose that there is a piecewise smooth curve of length $\ell$ joining $x$ and $y$. Then by using the bounded derivative assumption we have
$|u(x)-u(y)|\leq C\ell$ with some constant $C$. Now we have to construct such a curve with a reasonable length. If we can connect $x$ and $y$ by a straight line in $\Omega$, we have $\ell=|x-y|$. So we assume this is not the case. Without loss of generality, also assume that $\Omega$ is given by $x_n<f(x_1,\ldots,x_{n-1})$, with $f\in C^{0,\alpha}$. Let us denote by $x'$ and $y'$ the projections of $x$ and $y$, respectively, onto the plane $\{x_n=0\}$. Then we look at the function $f$ along the line segment $[x'y']$, and denote its minimum by $h$.
There are two cases: 1) $x_n>h$ and $y_n>h$, and 2) $x_n\leq h$ or $y_n\leq h$. We will only treat the case 1), the other case being easier. If we now redefine $x'$ and $y'$ to be the projections of $x$ and $y$, respectively, onto the plane $\{x_n=h-\delta\}$ with some small $\delta>0$, then the curve $L=xx'y'y$ will lie in $\Omega$. The length of this curve can be estimated as
$$
\begin{split}
\ell
&\leq|x-x'|+|x'-y'|+|y-y'|\leq f(x')-h+|x'-y'|+f(y')-h+2\delta\\
&\leq C_1|x'-y'|^\alpha+|x'-y'|+C_1|x'-y'|^\alpha+2\delta\\
&\leq C_2|x-y|^\alpha+2\delta,
\end{split}
$$
leading to
$$
|u(x)-u(y)|\leq CC_2|x-y|^\alpha + 2C\delta.
$$
As $\delta>0$ was arbitrary, we conclude that
$$
|u(x)-u(y)|\leq CC_2|x-y|^\alpha.
$$
Where you went wrong has been answered in a comment.
The result is not that hard. Your proof that a sequence in the unit ball of $C^{0,\alpha}$ must have a uniformly convergent subsequence is correct. So we may as well assume that $||f_n||_\alpha\le1$ and $f_n\to0$ uniformly, and we need to show that $||f_n||_\beta\to0$.
Let $\epsilon>0$. For every $n$ and every $x,y$ we have
$$|f_n(x)-f_n(y)|\le|x-y|^\alpha=|x-y|^{\alpha-\beta}|x-y|^\beta.$$Choose $\delta>0$ so that $\delta^{\alpha-\beta}=\epsilon$. Then the above shows that $$|f_n(x)-f_n(y)|\le\epsilon|x-y|^\beta\quad(|x-y|\le\delta).$$On the other hand, if $n$ is large enough then $2||f_n||_\infty/\delta^\beta<\epsilon$, so that $$|f_n(x)-f_n(y)|\le2||f_n||_\infty\le\frac{2||f_n||_\infty}{\delta^\beta}|x-y|^\beta<\epsilon|x-y|^\beta\quad(|x-y|>\delta).$$
(Similarly if $f_n\to g$ uniformly...)
Best Answer
It's Lemma 6.33 in Elliptic Partial Differential Equations of Second Order by Gilbarg and Trudinger. A proof is also presented by Ngô Quốc Anh on his blog. It's quite simple. Take a bounded family of functions in $C^\beta(\Omega)$. Extend them continuously to $\overline{\Omega}$ (uniform continuity allows that). Apply the Ascoli-Arzelà theorem to extract a uniformly convergent subsequence. Since $$\|u_m-u_n\|_{C^\alpha} \le \|u_m-u_n\|_{C^\beta}^{\alpha/\beta} (\sup |u_m-u_n|)^{1-\alpha/\beta} \tag1$$ it follows that $(u_m)$ is Cauchy in $C^\alpha$.
A parabolic Hölder space is a Hölder space with respect to a particular metric on $\mathbb R^n\times\mathbb R$, like $\rho((x,t),(x',t'))=|x-x'|+|t-t'|^{1/2}$. The above argument applies irrespective of the metric. All that matters is that the domain has a compact closure, so that the Ascoli-Arzelà theorem can be applied.