 Solution

First prove nowhere monotonicity of Brownian motion. On a fixed interval $[a,b]$ with rational endpoints and $a<b$, the $2^m$ increments over its equal subdivision are independent centered normal random variables. If the Brownian motion path were nondecreasing, all these increments would be nonnegative, an event with probability $2^{-2^m}$. Letting $m\to \infty$ gives probability zero. The same argument excludes nonincreasing paths. A countable union over rational intervals shows that, almost surely, no nontrivial interval supports a monotone function restriction of the path, since every such interval contains one with rational endpoints.

Work on this event and on the event of continuous paths. Inside any open interval $I\subset (0,\infty )$, choose two separated smaller intervals, the first to the left of the second. The first contains $r<s$ with $B_r<B_s$, because its restriction is not nonincreasing. The second contains $u<v$ with $B_u>B_v$, because its restriction is not nondecreasing. Thus $r<s<u<v$, all in $I$.

By the extreme value theorem, the path attains its maximum on $[r,v]$. This value exceeds $B_r$, since it is at least $B_s$, and exceeds $B_v$, since it is at least $B_u$. A maximizing time therefore lies in $(r,v)$ and is a local maximum of Brownian motion. Every open interval in the half-line contains a positive-time interval of this kind. Consequently the set of local maxima of Brownian motion is dense in $[0,\infty )$ almost surely.