= Solution
A continuous adapted process $M$ is a <local martingale> if there are <stopping times> $\tau_n\uparrow\infty$ almost surely such that each stopped process $M^{\tau_n}_t=M_{t\wedge\tau_n}$ is a <martingale>. In this definition $M_0$ is integrable. A continuous adapted process $A$ has <finite variation> if, almost surely, its paths have finite <total variation of a function> on each compact time interval:
$$
V_t(A)=\sup_{0=t_0<\cdots<t_r=t}\sum_{j=1}^r|A_{t_j}-A_{t_{j-1}}|<\infty.
$$
The permitted fact that this <total-variation process> is continuous and adapted makes its level-hitting times stopping times.
Suppose $M=A$, and put $N=M-M_0$. Define
$$
\sigma_n=\tau_n\wedge n\wedge\inf\{t:V_t(N)\geq n\}.
$$
These times increase to infinity almost surely. The <optional stopping theorem> applied to the localized martingale shows that $N^{\sigma_n}$ is a martingale; it is also bounded in absolute value by $n$, with total variation at most $n$. Fix $n,t$ and write $L=N^{\sigma_n}$. For a deterministic <partition of an interval> $0=t_0<\cdots<t_r=t$, orthogonality of square-integrable martingale increments gives
$$
\mathbb E L_t^2=\sum_j\mathbb E(L_{t_j}-L_{t_{j-1}})^2.
$$
Pathwise the sum on the right is bounded by $n\max_j|L_{t_j}-L_{t_{j-1}}|$. For partitions with mesh tending to zero, continuity makes this converge to zero. It is bounded by $2n^2$, so <dominated convergence> gives $\mathbb E L_t^2=0$. First use rational $t$, then continuity, to obtain $L\equiv0$ almost surely. Letting $n\to\infty$ proves
$$
\boxed{M_t=M_0\quad\text{for all }t\geq0\text{ simultaneously, almost surely}.}
$$
This proves <continuous finite-variation local martingale is constant> without assuming integrability of the unstopped total variation. Continuity matters: compensated jump martingales can have finite variation without being constant.
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