Proving the Harnack Inequality for Harmonic Functions I am putting together a proof for the Harnack inequality for harmonic functions defined on a balls. I would like to do this using the mean value property of Harmonic functions. 
I am thinking of doing this. Let us fix $x_0$, $y_0$ in some ball of radius $r$ centered around the origin. Now, we will construct two sub-balls around each of these balls. Specifically, let us take some arbitrarily small $\epsilon$ ball around $x_0$. Now, if we take $\epsilon$ small enough we can contain this sub-ball in a ball $B_{r_{y_0}} (y_0)$ around $y_0$ such that the ball around $y_0$ is a subset of our original ball $B_r$ on which the function is defined/harmonic. The mean value property then tells us that:
$$
C(\epsilon) \cdot u (x_0)  = \int_{B_{\epsilon}(x_0)} u(x) \mathrm{d}x
$$
and moreover:
$$
C(r_y) \cdot u(y_0) = \int_{B_{r_{y}}(y_0)} u(y) \mathrm{d}y
$$
Moreover, because the domain of the $y$ integral contains the domain of the $x$ integral:
$$
C(r_y) \cdot u(y_0) = \int_{B_{r_{y}}(y_0)} u(y) \mathrm{d}y \geq \int_{B_{\epsilon}(x_0)} u(x) \mathrm{d}x = C(\epsilon) \cdot u (x_0) 
$$
Hence:
$$
\frac{C(r_y)}{C(\epsilon)} u(y_0) \geq u(x_0)
$$
Now, my only issue with the above proof is dependencies. Specifically, it seems like my choice of constants determined is a function of the initial points I choose, which does not seem to be correct (through the general statement of the inequality). Could someone please help me verify if this is correct?
 A: 
my choice of constants determined is a function of the initial points I choose

This is inevitable. The constant in Harnack's inequality depends on the points involved. Usually it is stated as "for each compact subset $K$ of the domain there is a constant $C_K$ such that $u(x)/u(y)\le K$ for every $x, y\in K$ (where $u$ is a positive harmonic function)" but this doesn't really change anything: one has to take some $K$ that contains the given $x, y$.  
More importantly, your proof breaks down at this step:

we can contain this sub-ball in a ball $B_{r_{y_0}}(y_0)$ around $y_0$ such that the ball around $y_0$ is a subset of our original ball  

This cannot be done in general. For example, $x_0$ and $y_0$ could be these red dots: there is no ball centered at $y_0$ that is contained in the original ball $B$ and contains $x_0$.  

However, the argument can be repaired. It is valid when $|x_0-y_0| \le \frac12 \operatorname{dist}(y_0, \partial B)$, because one has $B(x_0, r)\subset B(y_0, 2r)\subset B$ where $r = |x_0-y_0|$. 
For general points, connect $x_0$ to $y_0$ by a line segment (in a general domain, use some curve) and place a finite sequence of points on that segment, locating them close enough to each other so that above applies.  
