= Solution
Use the representation from part (a), and put $\Delta_jM=M_{T_{j+1}}-M_{T_j}$, $U_j=h_j\Delta_jM$ and $N=H\cdot M$. The <optional stopping theorem> in the form needed here says that for a <uniformly integrable martingale> $L$ and <stopping times> $\sigma\leq\tau$, including terminal infinite times, $\mathbb E[L_\tau\mid\mathcal F_\sigma]=L_\sigma$. An <L2-bounded continuous martingale> is <uniformly integrable> and has the required terminal value.
Consequently $\mathbb E[\Delta_jM\mid\mathcal F_{T_j}]=0$. Conditioning on $\mathcal F_t$ and distinguishing whether $T_j\leq t$ gives
$$
\mathbb E[U_j\mid\mathcal F_t]=h_j\bigl(M_{t\wedge T_{j+1}}-M_{t\wedge T_j}\bigr).
$$
Thus $N_t=\mathbb E[\sum_jU_j\mid\mathcal F_t]$ is a <martingale>. Its paths are continuous, since each stopped increment is continuous and zero until its left endpoint.
For $i<j$, $U_i$ is $\mathcal F_{T_j}$-measurable, so the same conditional identity yields $\mathbb E(U_iU_j)=0$. These <martingale-difference orthogonality> relations also hold for the unweighted increments. Therefore
$$
\begin{aligned}
\mathbb E N_\infty^2&=\sum_j\mathbb E\bigl(h_j^2(\Delta_jM)^2\bigr)\\
&\leq\|H\|_\infty^2\sum_j\mathbb E(\Delta_jM)^2
=\|H\|_\infty^2\mathbb E(M_\infty-M_0)^2.
\end{aligned}
$$
Here $N_\infty=\sum_jU_j\in L^2$. By conditional Jensen, $\mathbb E N_t^2\leq\mathbb E N_\infty^2$ for every $t$, so
$$
\boxed{H\cdot M\in\mathcal M_c^2,\qquad\mathbb E(H\cdot M)_\infty^2\leq\|H\|_\infty^2\mathbb E(M_\infty-M_0)^2.}
$$
This proves the norm bound without presupposing the <Itô isometry> requested next.
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