Let $X_t,\, t\geqslant 0,$ be a Brownian motion and consider the stopping times $T_a := \inf \{t \mid X_t = a\}$. Find the probability $\mathbb P\{T_{2}< T_{-1} < T_{3}\}$, for instance.
So we have two events $\{T_2 < T_{-1}\}$ and $\{T_{-1}< T_3\}$. Separately, the probabilities are clear. My intuition says that we can multiply the probabilites for the initial problem, but I'm not really satisfied with this intuitive mumbo-jumbo. What if there are some funny cases, when it doesn't work..
Yet, I have no idea how to formally explain this.
They are definitely not independent.
You didn't specify the starting value for the Brownian motion, but let's say it's 0. On the event $\{T_3 < T_{-1}\}$, where the Brownian motion reaches 3 before -1, it must have passed through the value 2 even earlier, by continuity. So we have $\{T_2 < T_{-1}\} \subset \{T_3 < T_{-1}\}$. As such, $\{T_2 < T_{-1}\}$ cannot be independent of $\{T_3 < T_{-1}\}$, unless they both had probability 0 or 1 which is not the case. So $\{T_2 < T_{-1}\}$ is also not independent of $\{T_3 < T_{-1}\}^c = \{T_{-1} < T_3\}$.
(This same argument works if the starting point is anything less than 2. If it's greater or equal to 2, then $P(T_2 < T_{-1}) = 1$ and the events are trivially independent.)
You do end up multiplying probabilities, but not because of independence per se. The strong Markov property yields a certain conditional independence statement, if you like, but you have to be very careful to frame it the right way.