Let $(M_n)_{n\geq0}$ be a non-negative martingale with filtration $(\mathcal{F}_n)_{n\geq0}$. Set $$T=\min\{n\geq0:M_n=0\}.$$ Show that $M_n=0$ for all $n\geq T$ almost surely.
As $(M_n)_{n\geq0}$ is non-negative, it will be enough to show that the expectation for all such $n\geq T$ is zero. But there is no necessary reason that the martingale is uniformly integrable, nor that the stopping time is bounded, so I can't use the Optional Stopping Theorem, and so I'm not sure how to show this. Any hints or advice would be greatly appreciated!
As you observed, it suffices to check that $\mathbb E\left[M_n\mathbf{1}_{n\geqslant T}\right]=0$, since $M_n\geqslant 0$ will imply that $M_n\mathbb{1}_{n\geqslant T}=0$ almost surely.
To do so, we decompose $\mathbf{1}_{n\geqslant T}$ as the sum of $\mathbf{1}_{T=k}$ in order to get $$ \mathbb E\left[M_n\mathbf{1}_{n\geqslant T}\right]=\sum_{k=0}^n\mathbb E\left[M_n\mathbf{1}_{T=k}\right] $$ and to reduce the problem to show that $\mathbb E\left[M_n\mathbf{1}_{T=k}\right]=0$ for each $n\geqslant k$.
Denote by $\left(\mathcal F_n\right)_{n\geqslant -1}$ the filtration associated to the martingale $\left(M_n\right)_{n\geqslant 0}$. We notice that $\{T=k\}$ belongs to $\mathcal F_k$, as $\{T=k\}=\{M_k=0\}\cap\bigcap_{j=0}^{k-1}\{M_{j}>0\}$ if $k\geqslant 1$ and $\{T=0\}=\{M_0=0\}$. As a consequence, $$ \mathbb E\left[M_n\mathbf{1}_{T=k}\right]=\mathbb E\left[\mathbb{E}\left[M_n\mathbf{1}_{T=k}\mid\mathcal F_k\right]\right]= \mathbb E\left[\mathbf{1}_{T=k}\mathbb{E}\left[M_n\mid\mathcal F_k\right]\right] $$ and using the martingale property [until here, we only used $\mathcal F_n$-measurability of $M_n$], we get $$ \mathbb E\left[M_n\mathbf{1}_{T=k}\right]= \mathbb E\left[\mathbf{1}_{T=k}M_k\right] $$ which is $0$ because $\mathbf{1}_{T=k}M_k=0$ almost surely.