By Caldrron-Zygmund inequality, you may obtain : $$ \|u\|_{W^{2,1+\delta}~~(B_{\frac{1}{4}})} \leq C\|u\|_{L^{1+\delta}~~(B_{\frac{1}{2}})} +C\|g\|_{L^{1+\delta}~~(B_{\frac{1}{2}})}.$$ Since the dimension of space is 2, $$ \|u\|_{L^{\infty}~~(B_{\frac{1}{4}})}\leq C\|u\|_{W^{2,1+\delta}~~(B_{\frac{1}{4}})}.$$ Using the $L^1$- theory, you can obtain $\|u_1\|_{L^{1+\delta} ~~(B_{\frac{1}{2}})}$ is bounded. By the mean value theorem or harnack inequality you can obtain $\|u_3\|_{L^{\infty}~~(B_{\frac{1}{2}})}\leq C$, combine with the fact that $\|u_2\|_{L^{\infty}~~(B_{\frac{1}{2}})}\leq C$ you may obtain $$\|u\|_{L^{1+\delta}~~(B_{\frac{1}{2}})} \leq C,$$ and finish the proof.

By Caldrron-Zygmund inequality, you may obtain : $$ \|u\|_{W^{2,1+\delta}~~(B_{\frac{1}{4}})} \leq C\|u\|_{L^{1+\delta}~~(B_{\frac{1}{2}})} +C\|g\|_{L^{1+\delta}~~(B_{\frac{1}{2}})}.$$ Since the dimension of space is 2, $$ \|u\|_{L^{\infty}~~(B_{\frac{1}{4}})}\leq C\|u\|_{W^{2,1+\delta}~~(B_{\frac{1}{4}})}.$$ Using the $L^1$- theory, see

you can obtain $\|u_1\|_{L^{1+\delta} ~~(B_{\frac{1}{2}})}$ is bounded. By the mean value theorem or harnack inequality you can obtain $\|u_3\|_{L^{\infty}~~(B_{\frac{1}{2}})}\leq C$, combine with the fact that $\|u_2\|_{L^{\infty}~~(B_{\frac{1}{2}})}\leq C$ you may obtain $$\|u\|_{L^{1+\delta}~~(B_{\frac{1}{2}})} \leq C,$$ and finish the proof.