= 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 gives
$$
\mathbb E[U_t^2]=\sum_j\mathbb E[(U_{t_{j+1}}-U_{t_j})^2].
$$
Pathwise, however,
$$
\sum_j(U_{t_{j+1}}-U_{t_j})^2
\le\max_j|U_{t_{j+1}}-U_{t_j}|\sum_j|U_{t_{j+1}}-U_{t_j}|
\le n\max_j|U_{t_{j+1}}-U_{t_j}|\longrightarrow0,
$$
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 $U\equiv0$. Then let $n\to\infty$. This proves the <continuous finite-variation local martingale is constant> result in its zero-starting form:
$$
\boxed{X_t=0\text{ for every }t\ge0\text{ almost surely}.}
$$
The proof does not assume the uniqueness of <quadratic variation> that is to be established next.
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