Suppose that $(X_t, \mathscr{F}_t, 0 \le t < \infty)$ is a right-continuous submartingale and $S \le T$ are stopping times of $\mathscr{F}_t$. Then
$\{X_{T \wedge t}, \mathscr{F}_t : 0 \le t < \infty\}$ is a submartingale.
I have not found the answer to this problem anywhere and after some thought I think I proved this so I share my solution.
We need to show three things. First, $X_{T \wedge t}$ is $\mathscr{F}_t$ measurable for all $t \ge 0$. This follows since $X_{T \wedge t}$ is $\mathscr{F}_{T \wedge t}$ measurable.
Next, we need to show integrability. For this, take $T_n$, the decreasing sequence of stopping times that has $T$ as the infimum (c.f. Problem 1.2.24 of Karatzas and Shreve). Now for all $n \in \mathbb{N}$ and $t \ge 0$, since $T_n \wedge t \le t$, $T_n \wedge t$ will take finitely many values only by construction of $T_n$. Say $I$ is the index set for those finite values, then $$E|X_{T_n \wedge t}| = \sum_{i \in I} \int_{T_n \wedge t = i} |X_i| dP \le \sum_{i \in I} E|X_i| < \infty.$$
Now since $T_n \wedge t$ is decreasing, just as in the proof of Theorem 3.23 of Karatzas and Shreve, we can show that $\{X_{T_n \wedge t}, \mathscr{F}_{T_n \wedge t}\}$ is a backward submartingales. Specifically, for any $n$, it suffices to show that $E[X_{T_n \wedge t} | \mathscr{F}_{T_{n+1} \wedge t}] \ge X_{T_{n+1} \wedge t}$. But since we have $T_{n+1} \wedge t \le T_n \wedge t$ and both stopping times are bounded by $t$, we can use the discrete optional stopping theorem (9.3.4 from Chung) to conclude that we have the desired inequality. Now for any $n$, $0 \le T_n \wedge t \le t$, again by Theorem 9.3.4 of Chung, $\{X_0, X_{T_n \wedge t}\}$ forms a submartingale for each $n$, which gives $EX_0 \le E X_{T_n \wedge t}$ for all $n$. Hence we can use Theorem 3.11 (any backward submartingale $X_n$ with $\lim_n E(X_n) > -\infty$ is uniformly integrable) to conclude that $\{X_{T_n \wedge t}\}$ is uniformly integrable. Since the process is right continuous, we have $X_{T \wedge t} = \lim_n X_{T_n \wedge t}$ a.e., and it follows from uniform integrability that the limit is also in $L^1$, which implies that $X_{T \wedge t}$ is integrable.
Finally, we need to show that for $s \le t$ $$E[X_{T \wedge t} | \mathscr{F}_s] \ge X_{T \wedge s}.$$
For this, I do not know how to show it using any knowledge from before. However, looking forward and using Problem 3.26, we can easily prove this. For any two bounded stopping times $\sigma \le \tau$, it is clear that $T \wedge \sigma \le T \wedge \tau$ are bounded stopping times, too. Therefore,
$$E X^T_\sigma := EX_{T \wedge \sigma} \le E X_{T \wedge \tau} =: EX^T_\tau,$$ hence by 3.26, $X^T$ is a submartingale for $\mathscr{F}_t$.
Step 1: $X_{t \wedge T}$ is adapted. This is obvious since $X_{t \wedge T}$ is $\mathcal{F}_{t \wedge T}$-measurable and $\mathcal{F}_{t \wedge T} \subset \mathcal{F}_{t}$.
Step2: $X_{t \wedge T}$ is integrable.
$\mathbb{E} |X_{t \wedge T}| = -\mathbb{E}X_{t \wedge T} + 2\mathbb{E}X^+_{t \wedge T} \leq -\mathbb{E}X_{0} + 2\mathbb{E}X^+_{t} < \infty, \forall t$, using the submartingale inequality.
Step3: $\mathbb{E}(X_{t \wedge T} |\mathcal{F_{s}}) \geq X_{s \wedge T}, 0\leq s< t < \infty$
This is very similar to the proof in Theorem 3.22. Define the discrete $T_{n}$ as in Problem 1.2.24 of Karatzas and Shreve. Let $Y_{n} := X_{t \wedge T_{n}}$. Then $Y_{n}$ is a backward submartingale. $\{\mathbb{E}Y_{n}\}_{n=1}^{\infty}$ is decreasing and bounded below by $\mathbb{E}X_{0}$. By problem 3.11, $\{Y_{n}\}_{n=1}^{\infty}$ is uniformly integrable. We already have that $\mathbb{E}(X_{T_{n}\wedge t}|\mathcal{F}_{s \wedge T_{n}}) \geq X_{T_{n}\wedge s}$. Hence $\mathbb{E}(X_{T_{n}\wedge t}\mathbb{1}_{A_{n}}) \geq \mathbb{E}(X_{T_{n}\wedge s}\mathbb{1}_{A_{n}}), \forall A\in \mathcal{F}_{T_{n}\wedge s}$. Sending $n \uparrow \infty$ and by the previous argument that $\{Y_{n}\}_{n=1}^{\infty}$ is UI, we have that $\mathbb{E}(X_{T \wedge t}\mathbb{1}_{A}) \geq \mathbb{E}(X_{T\wedge s}\mathbb{1}_{A}),\forall A \in \mathcal{F}_{T\wedge s}$. On the event $\{T \geq s\}$, $\mathcal{F}_{T\wedge s} = \mathcal{F_{s}}$ and the result follows. On the event $\{T < s\}$, $\mathbb{E}(X_{t \wedge T} |\mathcal{F_{s}}) = X_{T} \leq X_{T\wedge s}$. We have proved the submartingale inequality.