Let $B=(B_t)_{t\ge 0}$ be a standard Brownian motion and $$\tau:=\inf\big\{t\ge 0:B_t\in\left\{a,b\right\}\big\}$$ for some $a<0<b$. I want to prove, that $\tau$ is almost surely finite.
Let $$\tau_x:=\inf\left\{t\ge 0:B_t=x\right\}\;\;\;\text{for }x\in\mathbb R\;.$$ Since $\tau=\tau_a\wedge\tau_b$, it's sufficient to show, that $\tau_x$ is almost surely finite.
How can we prove $\operatorname P\left[\tau_x<\infty\right]=1$ from the definition of Brownian motion and it's basic properties (like scaling invariance)?
I've seen some proofs which state, that $$\operatorname P\left[\tau_x\le t\right]\ge\operatorname P\big[\left|B_t\right|\ge x\big]\;\;\;\text{for all }t\ge 0\tag 1$$ and then concentrate on the right-hand side.
But how can we rigorously prove $(1)$, i.e. not only mention, that it's "clear" that $(1)$ holds by continuity of the paths?
Since $$B_0=0\;\;\;\text{and}\;\;\;B\text{ is continuous}$$ almost surely, the following holds under $\operatorname P$: Let $T\ge 0$, $$m:=B_0\wedge B_T\;\;\;\text{and}\;\;\;M:=B_0\vee B_T\;.$$ Since $$m\le 0\;\;\;\text{and}\;\;\;M\ge 0\;,$$ the intermediate value theorem yields $$[m,M]\subseteq B_{[0,T]}\;.$$ Thus, $$\left\{\tau\le T\right\}=\left\{a\in B_{[0,T]}\vee b\in B_{[0,T]}\right\}\supseteq\big\{a\in [m,M]\vee b\in [m,M]\big\}\supseteq\big\{|B_T|\ge c\big\}\;,$$ where $$c:=|a|\vee b\;.$$