Suppose we have a continuous function $f : [0, \infty) \rightarrow \mathbb{R}$ (in particular, $f(t) \equiv B_t(\omega)$ for $B$ = Brownian motion $\omega \in \Omega$ for some probability space $(\Omega, \mathcal{F}, P)$) such that $$\limsup_{t \rightarrow \infty}f(t) = \infty \quad \text{and} \quad \liminf_{t \rightarrow \infty} f(t) = - \infty $$ Is it true that the set of accumulation points for $f(t)$ as $t \rightarrow \infty$ is the extended real line?
I think this is true, and have tried to justify this as follows, but I'm not sure if the proof is valid:
Fix $a \in \mathbb{R}$ and choose $M$ such that $M > |a|$.
Since $\limsup f(t) = \infty$, $\forall t, \exists s_1 \ge t$ such that $f(s_1) > M$.
Similarly, $\forall t \ge 0, \exists s_2 \ge t$ such that $f(s_2) < -M$. By the continuity of $f$ and the intermediate value theorem, there will exist an $s(a, t)$ between $s_1$ and $s_2$ such that $f(s(a,t)) = a$. Since this holds for every $t$, there is a subsequence $\{s_n\}_{n \in \mathbb{N}}$ such that
(1) $s_n \xrightarrow{n \rightarrow \infty} \infty$ and
(2) $f(s_n) = a$ for each $n$ and so $f(s_n) \xrightarrow{n \rightarrow \infty} a$
Hence $a$ is an accumulation point. The $\limsup$ and $\liminf$ assumptions guarantee that $\pm \infty$ are also limit points, so we are done $\square$