The ratio $\frac{u(z_2)}{u(z_1)}$ for positive harmonic functions is uniformly bounded on compact sets I want to prove the following:

If $E$ is a compact set in a region $\Omega \subset \mathbb C$, prove that there exists a constant $M$, depending only on $E$ and $\Omega$, such that every positive harmonic function $u(z)$ in $\Omega$ satisfies $u(z_2) \leq M u(z_1)$ for any two points $z_1, z_2 \in E$.

This appears as an exercise in Ahlfors' Complex Analysis text, regarding Harnack's inequality:
If $u$ is positive and harmonic in the closed disk $\overline{\Delta}(z_0,R)$ then $$\frac{R-r}{R+r} \leq \frac{u(z)}{u(z_0)} \leq \frac{R+r}{R-r} $$ holds, whenever $|z-z_0|=r<R$.
My attempt:
Let $u:\Omega \to (0,\infty)$  be some harmonic function, and let $E \Subset \Omega$ be a compact subset. $\Omega$ 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 $E$ 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 $z,z' \in \Delta(z_n,\frac{1}{2} R_n)$ $$\frac{1}{3^2} \leq \frac{u(z)}{u(z_n)} \frac{u(z_n)}{u(z')}=\frac{u(z)}{u(z')} \leq 3^2 .$$
This shows that $\frac{u(z_2)}{u(z_1)}$ is bounded from above by 9, where $z_1,z_2$ lie in the same disk from the cover.
My difficulty is extending this idea to the case where $z_1,z_2 \in E$ lie in different disks.
Could you please help me? (I thought about taking intermediate steps $z_1 \to z_{i_1} \to \dots \to z_2$ such that each adjacent points lie in the same disk, but if $E$ is not connected this clearly fails).
Thanks!
 A: 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 $z^{(1)},z^{(2)} \in E$.
