Let $B_t$ be a standard Brownian motion. Use that by the Markov property $(B_{t+1/2}-B_{1/2})$ is also a standard Brownian motion independent of $B_{1/2}$. Use this to prove that $\mathbb{P}(\min_{1/2 \le t \le 1} B_t = 0) = 0$.
What I can think of is that we know that $B_t$ is standard normal, but I do not see how to apply this to this minimum. Could you please give me a hint?
Let $\xi:=\min_{0 \le s \le 1/2} B_{s+1/2}-B_{1/2}$. Then $$\mathbb{P}(\min_{1/2 \le t \le 1} B_t = 0) = \mathbb{P}( B_{1/2}+\xi = 0)=0 \,. $$ To see the last equality, condition on $\xi$, then use the independence of $B_{1/2}$ from $\xi$ and the known distribution of $B_{1/2}$.