I tried to solve this problem in the following way.
Suppose $\{B_t | t\in [0,1]\}$ is our Brownian motion. Define,
$$f_n(w) = \sum_{k=1}^{2^n} \bigg|B_{\frac{k}{2^n}}(w)-B_{\frac{k-1}{2^n}}(w)\bigg|.$$
Firstly, I showed that $f_n \leq f_{n+1}$ using triangle inequality.
We have to show that $f_n \xrightarrow{a.s} \infty$.
For that I want to prove first the following,
$\mathbb{P}(f_n\geq\alpha)\rightarrow 1\ \forall \alpha > 0....(*)$. Then from the definition of almost sure convergence we can conclude our desired result.
But I couldn't able to show the $(*)$. How to show that?
Let $Z$ be a standard normal, so that $E|Z|=\sqrt{2/\pi}$. $\;$Then $$E(f_n)= 2^{n/2}\cdot E|Z| = 2^{n/2} \cdot \sqrt{2/\pi} $$ and $$\text{Var}(f_n) \le 1 \,,$$ so by Chebyshev's inequality, $$P[f_n \le E(f_n)/2] \le \frac{4}{2^{n} \cdot 2/\pi}\,.$$