Probability of Brownian motion to have a zero in an interval

For any other initial x, construct a coupling between BMs started at x and 0, where the processes move in opposite directions until they meet (if they do), then they stick together afterwards. Then the answer is apparent.

There is an easy way to do it via calculation. Start by supposing that $x\geq 0$ Then we're asking for the probability that $x+W_t$ reaches $0$ in the interval $[a,b]$ where $W_t$ is a standard Brownian motion. This the same as $$\begin{eqnarray} P(\min_{a\leq t\leq b}x+W_t<0~and~\max_{a\leq t\leq b}x+W_t>0)=P(\min_{a\leq t\leq b}W_t<-x~and~\max_{a\leq t\leq b}W_t>-x) \end{eqnarray}$$ Then $$\begin{eqnarray} P(\min_{a\leq t\leq b}W_t<-x)&=& E[1\{\min_{a\leq t\leq b}W_t<-x \}| W_a]\\ &=& E[1\{\min_{a\leq t\leq b}W_{t-a}<-x-W_a \}| W_a]\\ &=& E[1\{\min_{0\leq t\leq b-a}W_{t}<-x-W_a \}| W_a]\\ &=& E[1\{-|W_{b-a}|<-x-W_a \}| W_a]\\ \end{eqnarray}$$ Where we used the fact that $W_t-W_a$ has the same law as $W_{t-a}$ and $\min_{0\leq t\leq b-a}W_{t}$ has the same law as $-|W_{b-a}|$ then we have

$$P(\min_{a\leq t\leq b}W_t<-x)=2\int_R \frac{\exp(-y^2/2a)}{\sqrt{2\pi a}}N(-y-x)dy $$

Where $N$ is the Gaussian cumulative. Now it is easy to see that the right hand side is decreasing as function of $x$. We treat then $x\leq 0$ in the same way by symmetry of the Brownian motion.