Let $\alpha \in (0,1]$, $\Omega \subset \mathbb{R}^n$ be an open bounded set and $u \in C_c^{0,\alpha}(\overline{\Omega})$. Let $\{\eta_\varepsilon\}$ be a family of standard (radially symmetric) smoothing kernels and set $u_\varepsilon := \eta_\epsilon \ast u$. Clearly, for every small enough $\varepsilon$, the corresponding functions $u_\varepsilon$ have compact support contained in $\Omega$. Moreover, each function $u_\varepsilon \in C^\infty(\Omega_\varepsilon)$, where $$\Omega_\varepsilon := \{ x \in \Omega : {\rm dist}(x, \partial \Omega) > \varepsilon \}.$$Notice that in our setting the Holder seminorm $$[u]_{C^{0,\alpha}_c(\overline\Omega)} := \sup_{\substack{x,y \in \overline{\Omega} \\ x \neq y}} \frac{|u(x)-u(y)|}{|x-y|^\alpha}$$ is actually a norm, topologically equivalent to the standard one, i.e.: $$\|u\|_{C^{0,\alpha}_c(\overline{\Omega})} := \|u\|_{C(\overline\Omega)} + [u]_{C^{0,\alpha}_c(\overline{\Omega})}.$$ Now, it is easy to see that $$[u_\varepsilon]_{C^{0,\alpha}_c(\overline\Omega)} \leq [u]_{C^{0,\alpha}_c(\overline{\Omega})}.$$ Indeed, for all $x, y \in \Omega$ with $x \neq y$, we have $$\begin{split}|u_\varepsilon(x) - u_\varepsilon(y)| &= \left| \int_{\mathbb{R}^n} \eta(z)(u(x-z) - u(y-z))\,dz\right| \\ &\leq \int_{\mathbb{R}^n} \eta_\varepsilon(z) |u(x-z)-u(y-z)|\,dz \\ &\leq [u]_{C^{0,\alpha}_c(\overline\Omega)} \int_{\mathbb{R}^n} \eta(z) |x-z-y+z|^\alpha \,dz \\ &= [u]_{C^{0,\alpha}_c(\overline\Omega)} |x-y|^\alpha \end{split}$$ as $\int_{\mathbb{R}^n} \eta_\varepsilon(z)\,dz = 1$ for each $\varepsilon > 0$, and hence the conclusion follows by dividing both sides by $|x-y|^\alpha$ and passing to the supremum over $x, y \in \overline{\Omega}$, $x \neq y$, in both sides.
Next, I would like to prove the following (qualified as "easy to check" in a paper I am reading)
Goal. In the same notation as before, there exist a constant $C > 0$ such that, for every $\varepsilon > 0$ small enoug and every $u \in C^{0,\alpha}_c(\overline{\Omega})$, it holds $[u_\varepsilon]_{C^{0,1}_c(\overline{\Omega})} \leq C \varepsilon^{\alpha-1}[u]_{C^{0,\alpha}_c(\overline\Omega)}$.
Although the result is reasonable, I did not succeed in adapting the previous argument to show this additional conclusion. Of course, I also tried to use the symmetry of the convolution, i.e., the fact that $$\int_{\mathbb{R}^n} \eta_\varepsilon(z)u(x-z)\,dz = \int_{\mathbb{R}^n} \eta_\varepsilon(x-z)u(z)$$ and the fact that each $\eta_\varepsilon$ is indeed Lipschitz, with Lipschitz constant of order $1/\varepsilon$ but I was not able to combine all these pieces of information in the right way. Any help is highly appreciated.