$\Omega$ open connected of $\mathbb{R}^N$ and $K\subset \Omega$ compact, then $c u(x) \le u(x')\le C u(x)$ for $u$ harmonic 
Let $\Omega$ be a connected open of $\mathbb{R}^n$ and $K\subset
 \Omega$ a compact. Show that there exists constants $c>0$ and $C>0$,
  depending only on $K$, such that if $u$ is an harmonic function in
  $\Omega$, $u\ge 0$, then
$$c u(x) \le u(x')\le C u(x), \forall x,x'\in K$$

My attempt:
Let $x\in K$ and $\{U_{\alpha}\}_{\alpha\in I}\subset \mathbb{R}^n$ open set
$K\subset \cup_{\alpha\in I}U_{\alpha}\implies$ there exists $U_{\alpha_1}, \cdots, U_{\alpha n}$ such that $K\subset \cup_{i=1}^n U_{\alpha i}$
which implies that there exists $U_{\alpha i}$ for some $i\in\{1,\cdots,n\}: x\in U_{\alpha i}$ which impies that $\exists r_i>0$ such that $B_{ri}(x)\subseteq U_{\alpha i}$
Let $\overline{r_i}:=\min\{dist(k,x), r_i\}$ then by Harnack's inequality
$$\left(\frac{\overline{r_i}-r}{\overline{r_i}+r}\right)^Nu(x)\le u(x') \le u(x)\left(\frac{\overline{r_i}+r}{\overline{r_i}-r}\right)^N,\ \forall x,x'\in B_{\overline{ri}}(x)\subseteq K, \forall 0<r\le \overline{ri}$$
but I don't know how to conclude by showing it works for all $x, x'$ and I also didn't use the fact that it is connected.
UPDATE: 
I've found something that is almost what I need here but it proves for $u$ positive, not $u\ge 0$. Can it be modified to prove what I want?
UPDATE:
Can somebody explain me this part?

If $x$ is fixed, then why $\frac{u(z)}{u(y)}$ is finite? Shouldn't it be $\frac{u(z)}{u(x)}$?
 A: Given a ball $B\Subset\Omega$, you know by Harnack's inequality that $\max_B u \leq C_B \min_B u$.
Given a compact set $K\Subset\Omega$, you have $K\subset B_1\cap\dots\cap B_n$ for some balls $B_i\Subset\Omega$. Now you want to "glue together" the information to obtain $\max_K u\leq C_K\min_K u$. This implies
$$
u(x') \leq \max_K u \leq C_K \min_K u \leq C_K u(x) \qquad \forall x,x'\in K,
$$
from which you get also $C_K^{-1} u(x)\leq u(x')$, of course.
Edit.
I now realize that what you are interested in might in fact be the actual "glueing". You do that by chaining. Given $x,x'\in K$, construct a sequence $x=x_0,x_1,x_2,\dots,x_m=x'$ of points in $K$ such that for every $i$ you have $x_i,x_{i+1}\in B_j$ for some $j$. Then
$$
u(x_0) \leq C_{B_{i(0)}} u(x_1) \leq C_{B_{i(0)}}C_{B_{i(1)}} u(x_2) \leq
\dots \leq C_{B_{i(0)}}\dots C_{B_{i(m-1)}} u(x_m)
\leq C_{B_1}\dots C_{B_n} u(x_m).
$$
This yields $C_K \leq C_{B_1}\dots C_{B_n}$.
Edit 2.
How to construct the chain?
Fix $x,x'\in K$. I claim that there is a sequence $B_{i(0)},\dots,B_{i(m)}$ taken from the balls above that are covering $K$ such that:


*

*each ball is taken at most once

*$x\in B_{i(0)}$

*$x'\in B_{i(m)}$

*$B_{i(j)}\cap B_{i(j+1)}\neq\emptyset$
Given this, just take $x_j\in B_{i(j)}\cap B_{i(j+1)}$ and notice that it has the properties I stated before.
A: To update from $u>0$ to $u\geq0$.
Suppose that the Harnack inequality is true for $u>0$ harmonic. Pick an arbitrary harmonic function $u\geq0$ in $\Omega$. Then $u+\varepsilon$ is positive harmonic for any $\varepsilon>0$. Apply the Harnack inequality to $u+\varepsilon$ and then send $\varepsilon\to0$.
A: If you know the result holds for positive $u$ you can apply the maximum/minimum principle: if $u \ge 0$ and $u = 0$ at an interior point of $\Omega$ (assuming $\Omega$ is connected) then $u$ is identically zero. (Theorem 1.8 in the book you linked)
Thus unless $u$ is identically zero, $u$ attains a strictly positive minimum value on any compact $K \subset \Omega$.
