I am studying Brownian Motion from Durrett's Probability: Theory and Examples.
In Theorem 7.2.9, Durrett claimed that one-dimensional Brownian motion is recurrent. The complete statement of the theorem is:
Let $B_t$ be a one-dimensional Brownian motion and let $A = \cap_n\{B_t = 0$ for some $t \geq n\}$. Then $P_x(A) = 1$ for all $x$.
The reason is because with probability $1$, $\lim \sup_{t \rightarrow \infty} B_t/\sqrt{t} = \infty$ and $\lim \inf_{t \rightarrow \infty} B_t/\sqrt{t} = -\infty$ for any one dimensional Brownian motion $B_t$ starting at $0$, the translation invariance property, and the continuity of Brownian paths.
I can understand the translation invariance property part; its role is to offset the $x$ in $P_x(A)$ (please correct me if I'm wrong), but the remaining part does not seem very obvious to me.
Question: Why does $\lim \sup_{t \rightarrow \infty} B_t/\sqrt{t} = \infty$, $\lim \inf_{t \rightarrow \infty} B_t/\sqrt{t} = -\infty$, and continuity implies $P_x(A) = 1$ for all $x$?