= Solution
First justify optional sampling for the square-minus-bracket process. The <Doob L2 maximal inequality> gives $\mathbb E\sup_{t\geq0}|M_t|^2\leq4\mathbb E|M_\infty|^2<\infty$. Stop the <local martingale> $M^2-[M]$ using a localizing sequence and bounded bracket levels. The stopped expectation identity, followed by Fatou, bounds $\mathbb E[M]_\infty$ by the expected squared path supremum. Hence $M^2-[M]$ has an integrable absolute supremum, and localization with conditional dominated convergence makes it a <uniformly integrable martingale>. The <optional stopping theorem> stated in part (b) is therefore applicable to both $M$ and $M^2-[M]$.
For ordered <stopping times> $\sigma\leq\tau$, expand the squared increment and condition on $\mathcal F_\sigma$. The mean increment is zero, so
$$
\mathbb E[(M_\tau-M_\sigma)^2\mid\mathcal F_\sigma]=\mathbb E[M_\tau^2-M_\sigma^2\mid\mathcal F_\sigma]=\mathbb E([M]_\tau-[M]_\sigma\mid\mathcal F_\sigma).
$$
This is the <conditional bracket isometry for stopped martingale increments>. Apply it with $\sigma=T_j$, $\tau=T_{j+1}$ and multiply by the bounded $\mathcal F_{T_j}$-measurable variable $h_j^2$. The orthogonality proved in part (b) gives the <Itô isometry>
$$
\boxed{\mathbb E(H\cdot M)_\infty^2=\sum_j\mathbb E\bigl[h_j^2([M]_{T_{j+1}}-[M]_{T_j})\bigr]=\mathbb E\int_0^\infty H_s^2\,d[M]_s.}
$$
To identify the whole bracket process, let $N=H\cdot M$ and $C_t=\int_0^tH_s^2\,d[M]_s$. For deterministic $s<t$ and $E\in\mathcal F_s$, apply the proved isometry to the simple integrand $\mathbf1_E\mathbf1_{(s,t]}H$. It yields
$$
\mathbb E[\mathbf1_E(N_t-N_s)^2]=\mathbb E[\mathbf1_E(C_t-C_s)].
$$
Thus $\mathbb E[(N_t-N_s)^2\mid\mathcal F_s]=\mathbb E[C_t-C_s\mid\mathcal F_s]$. Since $N$ is a <martingale>, the cross term $2N_s(N_t-N_s)$ has conditional mean zero. Consequently $N^2-C$ is a <martingale>. The process $C$ is continuous, adapted, nondecreasing and starts at zero, so uniqueness in the definition of <quadratic variation> proves
$$
\boxed{[H\cdot M]_t=\int_0^tH_s^2\,d[M]_s=(H^2\cdot[M])_t.}
$$
This deduction uses only the proved elementary isometry, not a previously assumed formula for the <quadratic variation> of <stochastic integrals>.
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