Let the dimension $d \ge 3$ and $(B(t))_{t\ge 0}$ be a $d$-dimensional Brownian motion starting at $0$ and $x \in \mathbb{R}^d$. I am looking at the proof of the following result from Rene Schilling's Brownian Motion and Stochastic Calculus.
If $d \ge 3$, Brownian motion is transient, i.e. $P^x ( \lim_{t\to \infty} |B(t)|=\infty)=1$.
Proof. By the strong Markov property, and the results on first return time, we find for all $r < R$,
$$P^0(\exists t \ge \tau_{B^c(0,R)}:B(t) \in B(0,r)) = E^0 (P^{B(\tau_{B^c(0,R)})}(\tau_{B(0,r)}<\infty))=(\frac{r}{R})^{d-2}.$$
Take $r=\sqrt{R}$ and let $R\to \infty$. This shows that the return probability of $(B(t))_{t\ge 0}$ to any compact neighborhood of $0$ is zero, i.e. $\lim_{t\to \infty}|B(t)|=\infty$.
I can't understand the final line of the proof. How does showing that $\lim_{R \to \infty} P^0(\exists t \ge \tau_{B^c(0,R)}:B(t) \in B(0,\sqrt{R}))=0$ imply that the return probability of $(B(t))$ to any compact neighborhood of $0$ is $0$?
Take $R_0$ such that the compact $K$ in question is contained in $B(0,\sqrt{R_0})$. For any $R\ge R_0$, the exit time $\tau_{B^c(0,R)}$ is finite almost surely. Therefore, the probability that $B$ will return to $K$ at arbitrarily large times does not exceed the probability that it will return to $K$ after exiting $B(0,R)$. In turn, the latter probability does not exceed $\left(\frac1{\sqrt{R}}\right)^{d-2}$. Letting $R\to +\infty$, we get that the probability in question is zero.