= Solution
A <simple predictable process> can be written
$$
H_s=\sum_{j=0}^{m-1}\xi_j\mathbf1_{(t_j,t_{j+1}]}(s),\qquad 0=t_0<t_1<\cdots<t_m\le\infty,
$$
where each $\xi_j$ is bounded and $\mathcal F_{t_j}$-measurable. Define its integral to start at zero:
$$
(H\mathbin\cdot M)_t=\sum_{j=0}^{m-1}\xi_j(M_{t\wedge t_{j+1}}-M_{t\wedge t_j}).
$$
Refining the partition leaves this sum unchanged. Every summand is a <martingale>, by conditioning its future increment on the information at $t_j$. Bounded coefficients and the $L^2$ bound on $M$ make the integral an <L2-bounded continuous martingale>. Its terminal value is the same sum with $M_\infty$ substituted at any infinite endpoint; $L^2$ <martingale> convergence supplies this limit.
We need the conditional square identity, rather than assuming <independent> increments. The <Doob L2 maximal inequality> gives $\mathbb E\sup_t|M_t|^2<\infty$. Localize the square identity for $M^2-[M]$ at level and bracket <stopping times>. The stopped identity and this maximal bound give $\mathbb E[M]_\infty<\infty$ by <monotone convergence>. Now
$$
\sup_t|M_t^2-[M]_t|\le\sup_t|M_t|^2+[M]_\infty
$$
is <integrable>, so <localization> passes to conditional <expectations> and $M^2-[M]$ is a <uniformly integrable martingale>. Expanding a squared increment and using $\mathbb E(M_t\mid\mathcal F_s)=M_s$ proves
$$
\mathbb E((M_t-M_s)^2\mid\mathcal F_s)=\mathbb E([M]_t-[M]_s\mid\mathcal F_s),\qquad s\le t\le\infty.
$$
The infinite endpoint follows from $L^2$ convergence and <integrability> of the bracket. The initial value $M_0$ need not be zero: the integral itself is zero-starting.
In the square of the terminal integral, all cross terms vanish. For $j<k$, the $j$th increment and both coefficients in its cross term are measurable at $t_k$, whereas the conditional mean of the $k$th increment there is zero. For the diagonal terms, the preceding conditional identity can be multiplied by the bounded $\xi_j^2$. Therefore
$$
\begin{aligned}
\mathbb E(H\mathbin\cdot M)_\infty^2
&=\sum_j\mathbb E\bigl[\xi_j^2(M_{t_{j+1}}-M_{t_j})^2\bigr]\\
&=\sum_j\mathbb E\bigl[\xi_j^2([M]_{t_{j+1}}-[M]_{t_j})\bigr]
=\boxed{\mathbb E(H^2\mathbin\cdot[M])_\infty}.
\end{aligned}
$$
This proves the <Itô isometry> for these simple integrands. Ordered stopping-time intervals also work, replacing the conditional identity by <optional stopping theorem>, but deterministic intervals suffice for the approximation requested here.
For that approximation, put $\mu(A)=\mathbb E\int_0^1\mathbf1_A(\omega,s)ds$ on the <predictable sigma-algebra>. This is a <finite measure>. That <sigma-algebra> is generated, apart from the zero-time slice of measure zero, by rectangles $A\times(s,t]$ with $A\in\mathcal F_s$. The indicator of every such rectangle is a bounded <simple predictable process>.
Here is a density proof. Let $V$ be the closed linear span of these indicators in $L^2(\mu)$, and let $\mathcal D$ contain the <predictable> sets whose indicators belong to $V$. The whole space belongs to $\mathcal D$; complements stay in $\mathcal D$; and disjoint countable unions do also, because their finite partial indicator sums converge in $L^2$ under the <finite measure>. Thus $\mathcal D$ is a <Dynkin system> containing the generating rectangle <pi-system>. The <Pi-lambda theorem> makes it the entire <predictable sigma-algebra>. Bounded <predictable> simple functions therefore belong to $V$. Truncating any $L^2(\mu)$ function and then approximating its truncated range by finitely many levels shows that every such function belongs to $V$. A finite sum of rectangle indicators can be rewritten on one common deterministic time partition, with the coefficient on each interval measurable at its left endpoint; it is exactly an allowed simple process.
Consequently choose $H^n$ of this form with $\|H^n-H\|_{L^2(\mu)}<1/n$. This proves
$$
\boxed{\mathbb E\int_0^1|H_s^n-H_s|^2ds\longrightarrow0.}
$$
It is <density of simple predictable processes for finite measures>, with a proof of the actual approximation step.
Finally apply the preceding isometry to $M_t=B_{t\wedge1}$, whose bracket is $t\wedge1$. The elementary integrals $I_n=\int_0^1H_s^n dB_s$ satisfy
$$
\mathbb E|I_n-I_m|^2=\mathbb E\int_0^1|H_s^n-H_s^m|^2ds\longrightarrow0.
$$
\b[Define the <Itô integral> as the $L^2$ limit of these elementary integrals.] The same identity shows <independence> of the approximating sequence, linearity, mean zero, and $\mathbb E(\int_0^1H_s dB_s)^2=\mathbb E\int_0^1H_s^2ds$. Applying the Doob inequality to the difference processes also gives uniform-in-time $L^2$ convergence on $[0,1]$, with a continuous <martingale> version of the integral.
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