The proof for a stopped discrete-time martingale is shown as follows.
Let $M=(M_n)_{n\ge0}$ be a discrete-time martinglae w.r.t. the filtration $(\mathcal F_n)_{n\ge0}$, and let $M^T=(M_{n\land T})_{n\ge0}$ be the stopped martingale, where $T$ is a stopping time w.r.t. $(\mathcal F_n)_{n\ge0}$. Since:
We have $$\begin{align} M_{n\land T}&=1_{T\ge n+1}M_n+1_{T\le n}M_T\\ &=1_{T\ge n+1}M_n+ \sum_{k=1}^n 1_{T=k}M_k\\ \end{align}$$ where $1_{T\ge n+1}, M_n, 1_{T=k}$, and $M_k$ are all $\mathcal F_n-measurable$, hence $M_{n\land T}$ is $\mathcal F_n-measurable$;
${\Bbb E}|M_{n\land T}| \le \underbrace{{\Bbb E}|M_1|+…+{\Bbb E}|M_n|}_{{\Bbb E}|M_i| \text{ is integrable, }\forall i\ge0}<\infty$, i.e., $M_{n\land T}$ is integrable $\forall n \ge 0$;
We have $$\begin{align} {\Bbb E}(M_{n+1 \land T}|{\mathcal F_n}) &= {\Bbb E}(M_{n \land T}+1_{T \ge n+1}(M_{n+1}-M_n)|{\mathcal F_n})\\ &=M_{n \land T}+1_{T \ge n+1}{\Bbb E}(M_{n+1}-M_n)\\ &=M_{n \land T}\\ \end{align}$$
Therefore, the stopped martingale satisfies the definition of a discrete martingale. Proof complete.
However, I am not able to extend these three parts of proof to a continuous version, because I cannot devide the time into separate spots with one next to another as the discrete version did. So I really wonder how to give a proof for a continuous-time martingale.