 Solution

Let $V_t$ be the total variation of $X$ on $[0,t]$, and set ${\tau }_n=\inf \{t:V_t\ge n\}$. Adaptation and continuity of $V$ make these stopping times; finite variation on every compact interval gives ${\tau }_n\uparrow \infty$. The stopped process $U=X^{{\tau }_n}$ satisfies $|U_t|\le n$ and has total variation at most $n$. It is a martingale by the bounded local martingale criterion, and is square-integrable.

Fix $t$ and take deterministic partitions $0=t_0<\cdots <t_m=t$ with mesh tending to zero. The martingale-difference orthogonality of the increments givesPathwise, however,by uniform continuity on $[0,t]$. These sums are bounded by $n^2$, so the dominated convergence theorem gives $\mathbb{E}{U}_t^2=0$. Hence $U_t=0$ almost surely. Take all rational $t$ on a common probability-one event and use continuity to obtain $U\equiv 0$. Then let $n\to \infty$. This proves the continuous finite-variation local martingale is constant result in its zero-starting form:The proof does not assume the uniqueness of quadratic variation that is to be established next.