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Continuity ensures that the maximum on $[0,1]$ is attained, so its first attainment time $\tau$ lies in $[0,1]$. To exclude the endpoint, use Brownian time reversal on a finite interval:This is a standard Brownian motion on this interval. Indeed, its increments are increments of $B$ on disjoint intervals taken in reverse order, hence are independent centered normal variables with the required variances, and its paths are continuous.

If $B_1$ is the maximum, then $R_s\ge 0$ for every $s$. But by the Brownian reflection principle and symmetry,the running maximum of $-R$ has the distribution of $|R_1|$, whose probability of being zero is zero. In particular the event $\tau =1$ has probability zero. Thus $\tau <1$ with probability one. Uniqueness of the maximizing time is not needed for this proof.