The ratio $\frac{u(z_2)}{u(z_1)}$ for positive harmonic functions is uniformly bounded on compact sets

Thanks to an answer provided by Daniel Fischer I think I can finish the proof:

I will start from the beginning:

Let $u:\Omega \to (0,\infty)$ be some positive harmonic function, and let $E \Subset \Omega$ be a compact subset. Using Daniel's answer we may find a compact and connected set $E_1 \supseteq E$. $\Omega$ can be covered by $ \left\{\Delta \left(z ,\frac{1}{2}R_z \right) \right\}_{z \in \Omega}$ where the radii $R_z>0$ are chosen such that the closed disks $\overline{\Delta}(z,R_z) $ remain in $\Omega$.

Since $E_1$ is compact it admits a finite subcover $\left\{ \Delta \left(z_n,\frac{1}{2} R_n \right) \right\}_{n=1}^N$.

Consider a disk $\Delta(z_n,\frac{1}{2} R_n)$ of the cover, Harnack's inequality implies that for any $z$ in that disk $$\frac{1}{3} \leq \frac{u(z)}{u(z_n)} \leq 3 .$$ Thus, for any two points $z,z' \in \Delta(z_n,\frac{1}{2} R_n)$ $$\frac{1}{3^2} \leq \frac{u(z)}{u(z_n)} \frac{u(z_n)}{u(z')}=\frac{u(z)}{u(z')} \leq 3^2 .$$

Let $z^{(1)},z^{(2)} \in E_1$. Since $E_1$ is connected, we have that the disks $\left\{ \Delta \left(z_n,\frac{1}{2} R_n \right) \right\}_{n=1}^N$ can ordered in such a way that $z^{(1)}$ lies in the first, $z^{(2)}$ lies in the last, and each disk intersects one of the previous ones (if they exist) in a non-empty set. Taking a representative $z_{i_k}$ from every intersection we can form a sequence $z^{(1)} \mapsto z_{i_1} \mapsto z_{i_2} \mapsto \dots \mapsto z^{(2)}$ using at most $N-1$ intermediate steps. Applying the previous result we find $$\frac{1}{3^{2N}} \leq \frac{u(z^{(2)})}{u(z^{(1)})} \leq 3^{2N} $$ for all $z^{(1)},z^{(2)} \in E_1$, and hence also for all $z^{(1)},z^{(2)} \in E$.