Harnack Inequality for uniformly elliptic PDE via constructing a singularity

Question

I am trying to prove a Harnack inequality for a nonnegative subsolution $u \in H^1(B_2)$ to the PDE $\text{div}(A Du) \ge 0$, where $A = A(x)$ is uniformly elliptic. The proof outline I am following is from a set of notes by a professor at my university, and the key step is the following inductive scheme:

Set $x_0$ to be a point such that $$u(x_0) = \sup_{B_{(0,1/2)}} u,$$ and choose $x_k$ inductively such that $x_{k+1}$ is such that $$u(x_{k+1}) = \sup_{B(x_k, r_k)} u$$ for $r_k$ sufficiently small to be chosen in a moment.

I have all of the steps except the following: suppose $$\frac{\text{sup}_{B_{0,1/4}} u}{ u(0)}$$ is sufficiently large, then we can choose a sequence $r_k$ such that $\sum r_k <1/2$ and a $\beta>1$ such that $u(x_{k+1}) \ge \beta u(x_k)$. That this would imply the result is immediate because it would contradict the boundedness of $u$. The preceding step, which I am led to believe is what implies the claim, is the following: $$u(x_{k+1}) \ge \frac{u(x_k) - cr_k^{-q} u(0)}{1-\theta}$$ where $c$, $q$ are absolute constants, and $1-\theta \ge \text{osc}_{B_1}u>0$ and $0<\theta \le \inf_{B_1} u$. Here $c,q>0$ are absolute constants.

I basically don't know what to do with this. Even if I assume the $\sup_{B_{1/4}}u / u(0)$ can get very large, the estimate (from the prior step) becomes useless as $r_k \to 0$, even after I replace $u(0)$ with $\sup_{B_{1/4}} u / N$ for large enough $N$. So it's unclear to me how to use it infinitely many times. I also suspect, but am not sure, that I should use the Nash-Digiorgi-Moser theorem here. Any suggestions or references would be much appreciated! I cannot find a similar proof anywhere, and given that I have provided the details for all of the other (numerous) steps, I would like to complete it.

Edit: Here is the full outline -- and yes, I know $u$ is Holder continuous on compact subsets $K \subset B_2$, with $[u]_\alpha \le C(K)\|u\|_\infty$.

1. Let $u: \Omega \to [0,\infty)$ be a supersolution, then show that if $B_{2R}(x_0) \subset \Omega$ and $r<R$, we have $$\inf_{B_R(x_0)} u \ge c \left(\frac{r}{R}\right)^q \inf_{B_r} u.$$

This follows from iteration of the weak Harnack inequality.

2. Let $u:B_2 \to [0,\infty)$ be a solution. Show that $$u(0) \ge cr_k^q \inf_{B_{r_k}(x_k)}u$$ for any $x_k \in B_{1/2}$ and $r_k$ small enough. This is simple enough.
3. As I defined the $x_{k+1}$ above, prove the inequality $$u(x_{k+1}) \ge \frac{u(x_k) - cr_k^{-q} u(0)}{1-\theta}.$$
4. Suppose $$\frac{\text{sup}_{B_{0,1/4}} u}{ u(0)}$$ is sufficiently large, then we can choose a sequence $r_k$ such that $\sum r_k <1/2$ and a $\beta>1$ such that $u(x_{k+1}) \ge \beta u(x_k)$. Conclude.

I have everything (including how to conclude), except for the choice of $r_k$ and $\beta$ in step $4$.