Supremum of Brownian motion almost surely not at endpoint

Question

Let $(B_t)$ be a standard $\mathbb R$-valued Brownian motion starting from 0. Let $$T = \inf\{t \in [0, 1] : B_t = \sup_{s \in [0, 1]} B_s\}. $$ I wish to show that $T < 1$ almost surely. Intuitively this should be true, with how jagged Brownian motion is I expect that almost surely for all $\epsilon > 0$ there exists $1 - \epsilon < t < \epsilon$ such that $B_t > B_1$. I'm unsure as to how to make this rigorous though.