 Solution

Put $t_j=j{2}^{-n}$ and ${\Delta }_j=X_{t_{j+1}}-X_{t_j}$. Telescoping $X_{t_{j+1}}^2-X_{t_j}^2=2X_{t_j}{\Delta }_j+{\Delta }_j^2$ givesThe martingale transform summands are orthogonal in $L^2$: for an earlier summand, conditioning on the sigma-algebra at the start of the later increment makes the cross expectation zero. HenceThe martingale increments themselves are also orthogonal. Since $X_0=0$, their variance sum equals $\mathbb{E}{X}_1^2\le C^2$. ThereforeOnly discrete martingale orthogonality is used here; no pre-existing quadratic variation calculation is needed.