Consider a stochastic process $X$ with right-continuous paths, which is adapted to a filtration $\{ \mathscr{F}_t \}$. Consider subset $\Gamma \in \mathscr{B}(\mathbb{R}^d)$ of the state space of the process, and define the hitting time as follows:
\begin{align*} H_{\Gamma} &: \Omega \to \mathbb{R} \cup \{\infty\} \\ H_{\Gamma}(\omega) &= \inf \{t \ge 0; X_t(\omega) \in \Gamma \} \\ \end{align*}
Where the infimum of the empty set is infinity.
If the set $\Gamma$ is open, show that $H_{\Gamma}$ is an optional time, such that $\{ H_{\Gamma} < t \} \in \mathscr{F}_t$.
The solution I have given below doesn't use the fact that $\Gamma$ is open or that the sample paths to $X$ are right-continuous. Is there a mistake or omission?
Consider the set:
\begin{align*} \bigcup\limits_{s \in \mathbb{Q}: 0 \le s < t} \{ X_s \in \Gamma \} \\ \end{align*}
This is the subset of $\Omega$ where for any $\omega$ in this subset there is some $s \in [0,t)$ such that $X_s(\omega) \in \Gamma$. Therefore $H_{\Gamma}(\omega) < t$ or $\omega \in \{ H_{\Gamma} < t \}$. This gives us:
\begin{align*} \bigcup\limits_{s \in \mathbb{Q}: 0 \le s < t} \{ X_s \in \Gamma \} \subset \{ H_{\Gamma} < t \} \\ \end{align*}
Now, for the converse direction, consider any $\omega \in \{ H_{\Gamma} < t \}$ such that $H_{\Gamma}(\omega) < t$. That means that for some $s < t$, we have $X_s(\omega) \in \Gamma$ or $\omega \in \{ X_s \in \Gamma \}$. This gives us:
\begin{align*} \{ H_{\Gamma} < t \} \subset \bigcup\limits_{s \in \mathbb{Q}: 0 \le s < t} \{ X_s \in \Gamma \} \\ \end{align*}
Combining the two directions we have:
\begin{align*} \{ H_{\Gamma} < t \} = \bigcup\limits_{s \in \mathbb{Q}: 0 \le s < t} \{ X_s \in \Gamma \} \\ \end{align*}
Since the expression on the right is a countable union of $\mathscr{F}_t$ measurable elements, the expression on the right is $\mathscr{F}_t$ measurable and therefore so is the equal expression on the left. Then we conclude that $H_{\Gamma}$ is an optional time.
Here is one counterexample to the statement $$\bigcup_{s\in\mathbb{Q}\cap[0,t)}\{\omega:X(\omega,s)\in \Gamma\}\neq \{\omega:H_\Gamma <t\}$$ in which right-continuity is not assumed.
Consider $\Omega=[0,1]$ with the Borel structure and the Lebesgue measure. Let $\Gamma=(1/2,\infty)$. Consider $X(\omega,t)=\mathbb{1}_{\Gamma}(\omega)\mathbb{1}_{\mathbb{R}\setminus \mathbb{Q}}(t)$. This function is in fact doubly-measurable (in $\omega$ and $t$) $$\bigcup_{s\in\mathbb{Q}\cap[0,t)}\{\omega:X(\omega,s)\in \Gamma\}=\emptyset$$ yet $$\{\omega:H_\Gamma <t\}=(1/2,1]$$
For general measurable $\Gamma$, it is possible to show, under some regularity assumptions in $\{\mathcal{F}_t: t\in\mathbb{R}\}$ and $X$, that $H_\Gamma$ is a stopping time. This however requires heavy machinery from the theory of Analytic sets.