Assume that $\{X_{t}, \mathcal{F_{t}}, 0 \leq t < \infty\}$ is a right-continuous submartingale and $S \leq T$ are stopping times of $\mathcal{F_{t}}$. Then $\mathbb{E}[X_{T\wedge t}|\mathcal{F_{S}}] \geq X_{S \wedge t}$ a.s. P, for every t $\geq 0$.
This question has been asked here. I have a different approach, but I'm not sure whether or not it works. In particular, I have not used the assumption that $X_{t}$ is right-continuous.
proof: From the optional stopping time theorem for bounded stopping times, $\mathbb{E}[X_{T\wedge t}|\mathcal{F_{S\wedge t}}] \geq X_{S \wedge t}$.
On $\{S \leq t\}$, $\mathbb{E}[X_{T\wedge t}|\mathcal{F_{S}}] = \mathbb{E}[X_{T\wedge t}|\mathcal{F_{S \wedge t}}] \geq X_{S \wedge t}$.
On $\{S > t\}$, $T \geq S > t$, $\mathbb{E}[X_{T\wedge t}|\mathcal{F_{S \wedge n}}] = \mathbb{E}[X_{t}|\mathcal{F_{S \wedge n}}] = X_{t} \geq X_{S \wedge t}$, where we choose $n > t$. We can send $n \uparrow \infty$ and by Lévy's theorem, $\mathbb{E}[X_{T\wedge t}|\mathcal{F_{S \wedge n}}] \rightarrow \mathbb{E}[X_{T\wedge t}|\mathcal{F_{S}}]$. Therefore, $\mathbb{E}[X_{T\wedge t}|\mathcal{F_{S}}] \geq X_{S \wedge t}$ as desired.