We want to prove the following implication : $$A\in \mathcal F_S \implies A\cap \{S\leq T\} \in \mathcal F_T$$
Since $A\in \mathcal F_S$, we have $A\in \mathcal F_\infty$ and $A\cap \{S\leq t\} \in \mathcal F_t$.
We will then show that $$A\cap \{S\leq T\} \in \mathcal F_T$$
Meaning :
$$ A\cap \{S\leq T\}\cap \{T\leq t\} \in \mathcal F_t $$
We have : $$ A\cap \{S\leq T\}\cap \{T\leq t\} = A\cap \{S\leq t\}\cap \{T\leq t\} \cap \{S\wedge t\leq T\wedge t\} $$
The ''chosen'' set $\{S\wedge t\leq T\wedge t\}$, according to the book, guarantees that the stopping time $S$ will be less or equal to $T$, and at the same time, the chosen set is in fact $\mathcal F_t$-measurable, which helps a lot in the proof
It's not very clear to me where it did come from.
I understand that
$$\{S \leq T\} \neq \{S\leq t\}\cap\{T \leq t\}, \hspace{1cm} \forall t\in \mathbb R_+$$
but the exact choice of $\{S\wedge t\leq T\wedge t\}$ is still not obvious.
$(T\wedge t \leq u)=(T\leq u\wedge t)\cup (T >t,t \leq u)$. Note that $(T >t,t \leq u)$ is empty if $u <t$ and it is $(T>t)$ if $u \geq t$. Conclude that $T\wedge t$ is meaurable on $(\Omega, \mathcal F_t)$. And so is $S\wedge t$ by the same argument. It follows that $T\wedge t-S\wedge t$ is also measurable on this sapce and $(T\wedge t-S\wedge t \geq 0)$ is measurable on $(\Omega, \mathcal F_t)$.