We are considering a Brownian Motion $(B_t)_t$ with values in $\mathbb{R} $ starting from $x$ defined on the stochastic basis:
$$(\Omega,\mathcal{E},(\mathcal{F}_t)_t,\mathbb{P}^x)$$
Then, let's define the stopping times:
$$\tau_{(a,b)}:=inf\{t>0:B_t\notin(a,b)\}$$
$$T_b=:=inf\{t>0:B_t=b\}$$
Lemma: $\mathbb{P}^a(\tau_{(a,b)}=0)=1$.
We want to conclude that $$\forall x\in\mathbb{R} \,\,\,\,\,\,\mathbb{P}^x(T_x=0)=1$$ The book give us this hint:
For the sake of simplicity take $x=0$. Let's define $T^-:=inf\{t>0:B_t<0\}$ and $T^+:=inf\{t>0:B_t>0\}$. Then prove, using the Lemma, that $\mathbb{P}^0(T^+=0)=1$ and $\mathbb{P}^0(T^-=0)=1$. So we can conclude that $\mathbb{P}^0(T_0=0)=1$.
My idea is to take in consideration $\tau_{(0,\infty)}$ and $\tau_{(-\infty,0)}$. But I can't prove the steps of the hint.
Any ideas?
Fix $\omega \in \Omega$. Since $[0,1] \ni t \mapsto B_t(\omega)$ is continuous, there exists $n \in \mathbb{N}$ such that
$$|B_t(\omega)| <n \tag{1}$$
for all $t \in [0,1]$. By the given Lemma, we have $\tau_{0,n}=0$ $\mathbb{P}^0$-a.s., i.e. there exists a sequence $(t_n)$, $t_n \to 0$ such that $B_{t_n}(\omega) \notin (0,n)$. From $(1)$, we see that $B_{t_n}(\omega) \leq 0$.
Using the symmetry, we find that there exists a sequence $(s_n)$, $s_n \downarrow 0$, such that $B_{s_n}(\omega) \geq 0$. Now the intermediate value theorem proves $T_0=0$ $\mathbb{P}^0$-a.s.