= Solution
The initial-value convention is essential. A <continuous finite-variation local martingale is constant>, so the conclusion is $M_t=M_0$ up to <indistinguishability of stochastic processes>. The stated zero conclusion uses the usual zero-start convention $M_0=0$; without it, $M_t\equiv1$ is a counterexample.
Put $N=M-M_0$. Let $V_t$ be its <total-variation process>, and stop when either $|N|$ or $V$ reaches $n$. The resulting process $N^{\tau_n}$ is a bounded <continuous local martingale>, hence a true <martingale>, and its total variation on every time interval is bounded by $n$. Fix $t$ and take deterministic partitions $0=t_0<\cdots<t_k=t$ whose mesh tends to zero. Orthogonality of <martingale> increments gives
$$
\mathbb E(N^{\tau_n}_t)^2=\sum_{j=1}^k\mathbb E\bigl(N^{\tau_n}_{t_j}-N^{\tau_n}_{t_{j-1}}\bigr)^2.
$$
Pathwise,
$$
\sum_j\bigl(N^{\tau_n}_{t_j}-N^{\tau_n}_{t_{j-1}}\bigr)^2
\leq\max_j|N^{\tau_n}_{t_j}-N^{\tau_n}_{t_{j-1}}|\,V_t(N^{\tau_n})\longrightarrow0
$$
by uniform continuity of a continuous path on $[0,t]$. The sums are bounded by $2n^2$, so the <dominated convergence theorem> makes their expectations tend to zero. Hence $N^{\tau_n}_t=0$ almost surely. Take the countable intersection over rational $t$ and integer $n$, use path continuity, and then let $n\to\infty$. The <stopping times> tend to infinity because the original path and its <finite variation> are bounded on every compact time interval. Thus
$$
\boxed{M_t=M_0\text{ for all }t\geq0\text{ outside one null event};\quad M_0=0\Longrightarrow M\equiv0.}
$$
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