The usual Harnack inequality says that $C^{-1}u(x) \leq \inf_{B_{\rho/2}(x)}u \leq \sup_{B_{\rho/2}(x)} u \leq Cu(x)$ for any $B_{\rho}(x) \subset B_1$ and $C$ universal depending on the ellipticity constants, etc. of $L$. Applying this with $x = 0$ and $\rho = 1$ gives
$$C^{-1}u(0) \leq \inf_{B_{1-2^{-1}}}u \leq \sup_{B_{1-2^{-1}}} u \leq Cu(0).$$
Applying it again for all $x \in \partial B_{1/2}$ and $\rho = 1/2$, and using the previous inequality, gives
$$C^{-2}u(0) \leq \inf_{B_{1-2^{-2}}}u \leq \sup_{B_{1-2^{-2}}}u \leq C^2u(0).$$
Proceeding inductively with $x \in \partial B_{1-2^{-k}}$ and $\rho = 2^{-k}$ gives
$$C^{-k}u(0) \leq \inf_{B_{1-2^{-k}}}u \leq \sup_{B_{1-2^{-k}}}u \leq C^ku(0),$$
so in particular
$$\sup_{B_{1-2^{-k}}}u \leq C^{2k} \inf_{B_{1-2^{-k}}}u.$$
For $r \in (0,\,1)$ choose $k \geq 1$ such that $1-2^{1-k} \leq r \leq 1-2^{-k}$, and let $p = 2\frac{\log C}{\log 2}$. Since $B_r \subset B_{1-2^{-k}}$ the previous inequality gives
$$\sup_{B_r} u \leq C^{2k} \inf_{B_r}u = 2^p(2^{1-k})^{-p} \inf_{B_r}u.$$
Since $1-r \leq 2^{1-k}$ the right side is bounded above by $2^p(1-r)^{-p}\inf_{B_r}u$, completing the proof.