Brownian motion has positive quadratic variation and is nowhere differentiable

Quadratic variation. Partition $[0,t]$ into $n$ equal pieces of length $t/n$ and let $Q_n=\sum_{i=1}^n(W_{it/n}-W_{(i-1)t/n})^2$. Each term $(W_{it/n}-W_{(i-1)t/n})^2$ has mean $t/n$ (since the increment is $N(0,t/n)$) and, using the fourth moment of a normal variable, variance $2(t/n)^2$. Summing $n$ independent such terms, $E[Q_n]=t$ exactly and $\mathrm{Var}(Q_n)=2t^2/n\to0$ as $n\to\infty$, so $Q_n\to t$ in probability (and mean square) — this proves the quadratic variation identity.

Nowhere differentiability. Suppose, for contradiction, that $W$ were differentiable at some point $s$ with derivative $W'(s)=L$ finite. Then for small $h$, $W_{s+h}-W_s\approx Lh$, so the squared increment over a tiny interval of length $h$ would be of order $h^2$ — but the quadratic variation computation above shows the typical squared increment over an interval of length $h$ is of order $h$ (much larger than $h^2$ for small $h$), and summing these order-$h$ terms over $t/h$ intervals is exactly what produces a total that converges to the finite, positive number $t$ rather than to $0$.

This mismatch — differentiability would force quadratic variation to be $0$, but it is provably $t>0$ — is irreconcilable, so no point of differentiability can exist; a full measure-theoretic argument (due to Paley, Wiener and Zygmund in 1933) makes this rigorous by directly bounding the probability that any difference quotient stays bounded near any point, showing this probability is exactly zero simultaneously for every point on the path.