= Solution
For $n>m$, define $r_j=\lfloor j2^{m-n}\rfloor2^{-m}$ and $h_j=X_{j2^{-n}}-X_{r_j}$. Each $h_j$ is measurable at time $j2^{-n}$, so the summands in the given difference formula are orthogonal <martingale transforms>. Therefore
$$
\mathbb E|M_1^{(n)}-M_1^{(m)}|^2=\mathbb E\sum_{j=1}^{2^n-1}h_j^2\left(X_{(j+1)2^{-n}}-X_{j2^{-n}}\right)^2.
$$
Let
$$
\omega_m=\sup\{|X_t-X_s|:s,t\in[0,1],\ |t-s|\le2^{-m}\}.
$$
Path continuity on the compact interval gives $\omega_m\to0$ almost surely, and $\omega_m\le2C$. Since $0\le j2^{-n}-r_j<2^{-m}$, $|h_j|\le\omega_m$. Thus, by the <Cauchy-Schwarz inequality> and part (b),
$$
\mathbb E|M_1^{(n)}-M_1^{(m)}|^2\le\mathbb E[\omega_m^2A_1^{(n)}]\le\sqrt{10}\,C^2(\mathbb E\omega_m^4)^{1/2}\longrightarrow0.
$$
The last step is the <dominated convergence theorem>. The bound is uniform in $n>m$, and the other ordering follows by symmetry. Hence \b[the terminal martingale transforms are Cauchy in $L^2$].
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