Let $(W_t)_{t \in \mathbb{R+}}$ be a Wiener process and define two random times for fixed $a>0$.
$S_a = \inf\{t > 0: W_t> a\}$
$T_a = \inf\{t > 0: W_t \geq a\}$
It is well known that $T_a$ is a stopping time with respect to the filtration generated by $(W_t)_{t \in \mathbb{R+}}$, while $S_a$ is not. Anyway the question is to show that $S_a=T_a$ almost surely.
I tried as follows; since $S_a \geq T_a$, we need to show that $P\{S_a > T_a\} = 0$. But $\{S_a > T_a \} = \bigcup_{q \in \mathbb{Q}} \{S_a \geq q\} \cap \{T_a < q\} $ and $\{S_a \geq q\}= \bigcap_{r \in [0,1], r \in \mathbb{Q}}\{W_{qr} \leq a\}$. But I'm stuck as it is sophisticated to express the event $\{T_a < q\}$ as above(a union or intersection of countable events). Any help would be greatly appreciated!
As you already noted, it suffices to show that $\mathbb{P}(S_a>T_a)=0$.
Step 1: Show the assertion for $a=0$, i.e. prove that $$S_0 = \inf\{t>0; W_t>0\}$$ satisfies $$\mathbb{P}(S_0>0)=0.$$
Possible approach: Apply Blumenthal's 0-1-law to show that $\mathbb{P}(S_0>0) \in \{0,1\}$. Use the symmetry of Brownian motion (i.e. the fact that $(-W_t)_{t \geq 0}$ is also a Brownian motion) to prove that $\mathbb{P}(S_0>0)=1$ cannot hold true. Hence $\mathbb{P}(S_0>0)=0$.
There are many other possibilites to prove this; it depends a lot on what you know about Brownian motion.
Step 2: Use the strong Markov property and Step 1 to prove the assertion for any $a>0$.