Let $(X_t)$ be a continuous nonnegative supermartingale and $T = \inf\{t\geq 0 \colon X_t = 0 \}$ then $X_t = 0$ for every $t\geq T$.
Idea of solution:
Since $T$ is stopping time, by Doob theorem: $$E(X_{T+q} 1_{T < \infty} | F_T) \leq X_T 1_{T < \infty} =0 $$ for every $q \in \mathbb{Q}$
Then taking expectation we have that $E(X_{T+q}1_{T < \infty}) \leq 0$ . Since $X_{T+q}1_{T < \infty} $ is positive $X_{T+q}1_{T < \infty} = 0$ a.e. (on a set $\Omega_q$ with $\mathbb{P}(\Omega_q)=1$). Taking $\Omega'=\cap_{q \in \mathbb{Q}^+} \Omega_q$ we will have that $$X_{T+t}1_{T < \infty} = 0$$ on $\Omega'$ for every $t\geq 0$.
The only problem is we are not allow to use Doob theorem since $T$ is not bounded and $X$ is not U.I. I try to use $T \wedge k $ to make the stoping time bounded but I couldn't take limit properly.
I will give simple proof for discrete time case. Someone else might extend it for contiunous time super-martingales.
First note that for any $A \in \mathcal{F}_k$ we have $$ E(M_t I_A)\le E(M_k I_A)$$
Now write $$ \begin{align} E(M_{T+x}) &= E(\lim_{n \rightarrow \infty}\sum_{k}^{n} M_{k+x} I_{T=k}) \\ &\le \liminf_{n\rightarrow \infty} \sum_{k}^{n} E(M_{k+x} I_{T=k}) \\ &\le \liminf_{n\rightarrow \infty} \sum_{k}^{n} E(M_{k} I_{T=k}) = 0 \end{align} $$ Here we used our first inequality with $A=I_{T=k}$, and Fatou’s lemma.
Hence, $0\le E(M_{T+k}) \le 0$, therefore $M_{T+k}=0$ almost surely for every k.