Almost sure non differentiability of Brownian Motion

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Problem: Let $t>0$, show that the standard Brownian motion is almost surely not differentiable a $t$

Now, through a Borel Cantelli argument I proved that, almost surely

$$\limsup_{\epsilon \rightarrow 0^+} \frac{|B_{t+\epsilon}-B_t|}{\epsilon} = \infty$$

Isn't this enough to prove non differentiability?

I ask this because I am given two more hints:

-Recall that by Blumenthal's 0-1 law, $\forall \epsilon > 0$ $B_{t+u}-B_{t} $ almost surely attains both a negative and a positive value for $ u\in [t,t+\epsilon]$.

-Conclude considering behaviours of $\limsup$ and $\liminf$ of $|B_{t+\epsilon}-B_t|/\epsilon$ as $\epsilon \rightarrow 0$

I also think that there shouldn't be a modulus sign in the last hint, it doesn't make much sense to me otherwise, what do you think?

Thanks.