Prove that if $S$ is an optional time and $T$ is a positive stopping time such that with $S \le T$ and $S < T$ on $\{S < \infty\}$, then $\mathscr{F}_{S+} \subset \mathscr{F}_T$.
The following is the given solution:
Consider any $A \in \mathscr{F}_{S+}$, which means that $\forall t \ge 0, A \cap \{S \le t\} \in \mathscr{F}_{t+}$. Then:
\begin{align*} A &= \left( \bigcup_{r \in \mathbb{Q}} [A \cap \{S < r < T \}] \right) \cup [A \cap \{S = \infty\}] \\ \end{align*}
For the left clause, $A \cap \{S < r < T\} = A \cap \{S < r \} \cap \{T > r\}$ is in $\mathscr{F}_T$ because $A \cap \{S < r\} \in \mathscr{F}_r$.
For the right clause, $A \cap \{S = \infty\} = [A \cap \{S = \infty\} ] \cap \{T = \infty\}$ is in $\mathscr{F}_T$. Then the conclusion follows.
All of that makes sense except I don't see how we can conclude that $A \cap \{S < r\} \in \mathscr{F}_r$
This follows from the definition of $\mathcal{F_{S+}}$. Recall that $A \cap \{S \leq t\} \in \mathcal{F_{t+}},\forall t\geq 0$. Now, $A \cap \{S < r\} = A \cap ( \cup_{n=1}^{\infty} \{S\leq r-\frac{1}{n}\}) = \cup_{n=1}^{\infty} (A \cap \{ S \leq r- \frac{1}{n} \})$. But $A \cap \{ S \leq r- \frac{1}{n} \} \in \mathcal{F_{(r-\frac{1}{n})+}} \subset \mathcal{F_{r}} \forall n$. Therefore, the union is in $\mathcal{F_{r}}$.