This is Problem 3.24 (ii) from Chapter 1 of Karatzas and Shreve.
Assume that $(X_t, \mathscr{F}_t: 0 \le t < \infty)$ is a right-continuous sub martingale and $S \le T $ are stopping times of $\mathscr{F}_t$. Then
$E[X_{T \wedge t} | \mathscr{F}_S] \ge X_{S \wedge t}$ a.s. $P$, for every $t \ge 0$.
We know that $E[X_{T \wedge t} | \mathscr{F}_{S \wedge t}] \ge X_{S \wedge t}$ from the optional sampling theorem for bounded stopping times. This means that
$$\int_A X_{T \wedge t} dP \ge \int_A X_{S \wedge t} dP$$ for all $A \in \mathscr{F}_{S \wedge t} = \mathscr{F}_S \cap \mathscr{F}_t$. Now we want to show the above inequality for $A \in \mathscr{F}_S$.
So take any $A \in \mathscr{F}_S$ that is not in $\mathscr{F}_t$. Then we have $A = (A \cap \{S \le t\}) \cup (A \cap \{S > t\})$. The former set is in $ \mathscr{F}_S \cap \mathscr{F}_t$ while the latter is only in $\mathscr{F}_S$. We have the inequality on the former set. But we have $\int_{A \cap \{S > t\}} X_{S \wedge t}dP= \int_{A \cap \{S > t\}} X_t dP$ and similarly $\int_{A \cap \{S > t\}} X_{T \wedge t}dP= \int_{A \cap \{S > t\}} X_t dP$ since $S \le T$ implies that on the set $S>t$ we also have $T > t$. Hence, if we sum the two integrals, we get the desired inequality for all $A \in \mathscr{F}_S$.