Question on the proof of Lemma 6.10 in Gilbarg/Trudinger

This lemma states that for boundary condition $$\varphi\in C^0(\partial B) \cap C^{2,\alpha}(T)$$ where $T$ is a portion of $\partial B$, the Dirichlet problem $$Lu = f$$ is solvable on ball $B$, with the assumptions $L$ is uniformly elliptic and its coefficients $$a^{ij},b^{i},c, f\in C^{\alpha}$$ where $c\leq 0$.
Its proof is to first extend $\varphi$ on to whole $B$ and approximate by smooth $\varphi_k$ in $C_0$-norm, and from the potential theory there is corresponding $u_k\in C^{2,\alpha}(\bar{B})$ solving boundary condition $\varphi_k$.

Question: So I know from Shauder estimate we could conclude $u_k$ converges in $C^2$, but the author then states, without futher explanation that

It follows from $$\vert \varphi_k - \varphi\vert_{0;B}\to 0\; \text{ and }\; \vert \varphi_k \vert_{2,\alpha,G} \leq C\vert \varphi\vert_{2,\alpha,G}$$ ($G$ is a small ball centered around $x_0\in T$), the Shauder estimate $$\vert u_k\vert_{2,\alpha,D} \leq C(\vert u_k\vert_{0,B} + \vert\varphi_k\vert_{2,\alpha,G} + \vert f \vert_{0,\alpha, B})$$ ($D$ is a smaller ball contained in $G$) and Arzela's theorem that $$u\in C^{2,\alpha}(\bar{D})$$

but I can only see $u\in C^2$, where does that $C^{2,\alpha}$ come from?

It's not exactly that GT are asserting convergence in $C^{2,\alpha}$ but rather that what they know at that point is sufficient to deduce that the limit $u$ belongs to $C^{2,\alpha}$. The idea is that once you have convergence in $C^2$ together with a uniform bound of the second derivatives in the Hölder space $C^{0,\alpha}$, you can pass to the limit in the Hölder bound to deduce that $u \in C^{2,\alpha}$.

Indeed, let's say that we know $u_k \to u$ in $C^2$ and that $$ \sup_{k} \sup_{|\beta|=2} \sup_{x\neq y}\frac{|\partial^\beta u_k(x) - \partial^\alpha u_k(y)|}{|x-y|^\alpha} \le M < \infty. $$ Unpacking this, for any fixed $k$ and $\beta$ with $|\beta|=2$ we have that $$ |\partial^\beta u_k(x) - \partial^\beta u_k(y)| \le M |x-y|^\alpha $$ (note we don't have to worry about $x\neq y$ once we rewrite in this form). Due to the convergence in $C^2$, we have that $\partial^\beta u_k \to \partial^\beta u$ pointwise, so we can pass to the limit in our bound above to see that
$$ |\partial^\beta u(x) - \partial^\beta u(y)| \le M |x-y|^\alpha. $$ This holds for all such $\beta$ and for all $x,y$, so we deduce that $$ [u]_{2,\alpha} = \sup_{|\beta|=2}\sup_{x\neq y} \frac{|\partial^\beta u(x) - \partial^\beta u(y)|}{|x-y|^\alpha} \le M. $$ Thus $u \in C^{2,\beta}$ and the bound $M$ passes to the limit as a bound for the limiting Hölder seminorm.