Variants of this question have been asked many times (e.g. here) but none of them touch on a particular step. I will first state the theorem and proof (from Protter's Stochastic Integration and Differential Equations} and then my attempt afterwards.
Theorem statement: Let $X_t$ be a adapted cadlag process and let $\Lambda$ be an open set. Then the hitting time, T, of $\Lambda$ for $X_t$ is a stopping time.
Proof from Protter: It suffices to show $\{T<t\} \in \mathcal{F}_t$, $0 \leq t \leq \infty$. We have, \begin{equation} \{T < t\} = \bigcup_{s\in \mathbb{Q} \cap [0, t)} \{X_s \in \Lambda\}. \end{equation} Then, $\{X_s \in \Lambda\} = X_s^{-1}(\Lambda) \in \mathcal{F}_s$ since $X_s$ is measurable and $\Lambda$ is open. Hence, $\{T < t\} \in \mathcal{F}_t$ and since $t$ was arbitrary the result follows.
Now, the part I do not completely follow is the equality: \begin{equation} \{T < t\} = \bigcup_{s\in \mathbb{Q} \cap [0, t)} \{X_s \in \Lambda\}. \end{equation} So I have tried to fill in the gaps, here is my proof of the above equality:
First,
\begin{equation} \{\omega:T(\omega)<t\} = \{\omega: \exists s<t \text{ such that } X_s(\omega) \in \Lambda\} =\bigcup_{s\in [0, t)} \{ \omega: X_s(\omega) \in \Lambda\}. \end{equation} So it suffices to show equality of $A :=\bigcup_{s\in [0, t)} \{ X_s \in \Lambda\}$ and $B:=\bigcup_{s\in \mathbb{Q} \cap [0, t)} \{X_s \in \Lambda\}$. Clearly $B \subset A$. To show $A \subset B$, take $\omega \in A$. Then $\exists s<t$ with $X_s(\omega) \in \Lambda$ (and we assume $s \not\in \mathbb{Q}$). Then there exists a sequence $\{s_n\}$ such that $s_n \searrow s$ through $\mathbb{Q}$. This implies, by right continuity of paths of $X_t$, that \begin{equation} X_s(\omega) = \lim_{n\rightarrow \infty}X_{s_n}(\omega) \end{equation} Since $\Lambda$ is open and $X_s(\omega) \in \Lambda$, then $s_N \in \mathbb{Q}$, $s_N < t$ can be chosen with $X_{s_N}(\omega) \in \Lambda$. This implies $\omega \in B$ and we are done.
Is my attempt at filling in the gaps correct?