 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, $M_t\equiv 1$ 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 givesPathwise,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 $N_t^{{\tau }_n}=0$ 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