= Solution
An <elementary predictable process with stopping-time intervals> has the form $H=\sum_{j=0}^{r-1}h_j\mathbf1_{(\tau_j,\tau_{j+1}]}$, where the <stopping times> are ordered and $h_j$ is bounded and $\mathcal F_{\tau_j}$-measurable. Deterministic endpoints give the usual <simple predictable process>. Define the zero-starting <stochastic integral> by
$$
\boxed{(H\cdot M)_t=\sum_{j=0}^{r-1}h_j(M_{t\wedge\tau_{j+1}}-M_{t\wedge\tau_j}).}
$$
Common refinement and telescoping show that this definition is independent of the representation. Use $M_\infty$ for an infinite endpoint. The <L2-bounded martingale convergence theorem> gives convergence of $M_t$ in $L^2$ to $M_\infty$ and $M_t=\mathbb E[M_\infty\mid\mathcal F_t]$.
Put $D_j=M_{\tau_{j+1}}-M_{\tau_j}$. The <optional sampling theorem> gives $\mathbb E[D_j\mid\mathcal F_{\tau_j}]=0$. For $j<k$, the variable $h_jD_jh_k$ is $\mathcal F_{\tau_k}$-measurable; hence $\mathbb E[h_jD_jh_kD_k]=0$. Applying the same theorem to the assumed <martingale> $M^2-[M]$ gives
$$
\mathbb E[D_j^2\mid\mathcal F_{\tau_j}]=\mathbb E[[M]_{\tau_{j+1}}-[M]_{\tau_j}\mid\mathcal F_{\tau_j}].
$$
Consequently the <Itô isometry> follows:
$$
\boxed{\mathbb E[(H\cdot M)_\infty^2]=\sum_j\mathbb E[h_j^2([M]_{\tau_{j+1}}-[M]_{\tau_j})]=\mathbb E\int_0^\infty H_s^2\,d[M]_s.}
$$
The unbounded <stopping times> cause no optional-sampling gap. The $L^2$ terminal representation makes $M$ <uniformly integrable> and makes the family of stopped $M^2$ <uniformly integrable>, by conditional Jensen applied to $M_\infty^2$. Also $\mathbb E[M]_\infty=\mathbb E M_\infty^2-\mathbb E M_0^2<\infty$, from <monotone convergence theorem> and the assumed square-compensator identity. Thus $M^2-[M]$ has the needed uniform integrability, and truncating <stopping times> and passing in $L^1$ is legitimate. Bounded coefficients are part of the simple-integrand convention; more general coefficients require the displayed square-integrability condition.
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