At some point I wanted to prove that if we have a continuous (trajectory) stochastic process $X_t$ (we consider its natural filtration) and $A$ a closed set, then $\tau(\omega)=\inf\left\{t | X_t(\omega)\in A\right\}$ is a stopping time.
My attempt is as follows :
$$ \left\{\tau(\omega)\leq t\right\} = \left\{ \omega | \inf\left\{ t | X_t(\omega)\in A\right\}\right\} = \left\{ \omega | \exists s\in [0,t] : X_s(\omega)\in A\right\} = \left\{ \omega | \exists s\in [0,t]\cap\mathbb{Q} : X_s(\omega)\in A\right\} $$
Where the last equality follows from the continuity of the trajectories of the process. Then, I would like to conclude by saying that we have a countable union of sets that are $\mathcal{F}_s$ measurable with $s\leq t$ and hence $\mathcal{F}_t$ measurable.
However, in my lecture notes it is not as simple as I do so I am wondering if I am missing something.
I believe the flaw is in the statement $$ \left\{ \omega | \exists s\in [0,t] : X_s(\omega)\in A\right\} = \left\{ \omega | \exists s\in [0,t]\cap\mathbb{Q} : X_s(\omega)\in A\right\}.$$
This is true if $A$ is open, but not if $A$ is closed. As an example, take $A := \{\sqrt{2}\}$ and $X_t(\omega) = t$ for all $\omega \in \Omega$. Then $\left\{ \omega | \exists s\in [0,2] : X_s(\omega)\in A\right\} = \Omega$, but $\left\{ \omega | \exists s\in [0,2]\cap\mathbb{Q} : X_s(\omega)\in A\right\} = \emptyset$.