= Solution
The correct <quadratic variation of a stochastic integral> identity is
$$
\boxed{[H\cdot M]_t=\int_0^t H_s^2\,d[M]_s=(H^2\cdot[M])_t.}
$$
The brackets around the final integrator are missing in the PDF. Without them the assertion is false: for $H=1$ and $M$ <Brownian motion>, the left side is $t$ while the printed right side is $B_t$.
For elementary <predictable> $H$, the integral is obtained by scaling each <martingale> increment by its known coefficient. <Quadratic variation> on each interval is therefore scaled by the square of that coefficient, giving the displayed formula. To pass to a general $L^2(M)$ integrand on $[0,T]$, take elementary $H_n\to H$ in the integrand norm and put $N_n=H_n\cdot M$, $N=H\cdot M$. The <Itô isometry> gives $\mathbb E(N_n-N)_T^2\to0$, hence $\mathbb E[N_n-N]_T\to0$. The bracket Cauchy-Schwarz bound
$$
|[N,N_n-N]_t|\le[N]_T^{1/2}[N_n-N]_T^{1/2}\qquad(t\le T)
$$
and polarization show that $[N_n]\to[N]$ uniformly in probability. On the other side, Cauchy-Schwarz for the <quadratic-variation measure> gives
$$
\mathbb E\int_0^T|H_n^2-H^2|\,d[M]\le\|H_n-H\|_{L^2(M;T)}\big(\|H_n\|_{L^2(M;T)}+\|H\|_{L^2(M;T)}\big)\longrightarrow0.
$$
Taking limits proves the formula in the square-integrable case. Stopping and pasting as in part b proves it for locally bounded <predictable> integrands and <continuous local martingales>.
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