= Solution
Fix a <L2-bounded continuous martingale> $M$. Its <quadratic-variation measure> on the <predictable sigma-algebra> is
$$
\nu_M(A)=\mathbb E\int_0^\infty\mathbf1_A(\omega,s)\,d[M]_s.
$$
It is finite, since $\nu_M(\Omega\times[0,\infty))=\mathbb E[M]_\infty=\mathbb E M_\infty^2-\mathbb E M_0^2$. The source <Hilbert space> is $L^2(M)=L^2(\Omega\times[0,\infty),\mathcal P,\nu_M)$: <predictable processes> with finite $\mathbb E\int H^2d[M]$, identified when equal $\nu_M$-almost everywhere. Its <norm> is $\|H\|_{L^2(M)}^2=\mathbb E\int H^2d[M]$.
The target $\mathcal M_0^2$ consists of <L2-bounded continuous martingales> starting at zero, identified up to <indistinguishability of stochastic processes>, with <norm> $\|K\|_{\mathcal M_0^2}=\|K_\infty\|_2$. It is a <Hilbert space>: terminal values belong to the closed <linear subspace> of $L^2(\mathcal F_\infty)$ whose <conditional expectation> given $\mathcal F_0$ is zero and whose associated <martingales> have continuous versions. Closure of that continuous-version subspace follows from the <Doob L2 maximal inequality> and an almost surely uniformly convergent subsequence.
The <Itô isometry> says that integration extends uniquely from <simple predictable processes> to a linear <isometry> $L^2(M)\longrightarrow\mathcal M_0^2$, with
$$
\boxed{\mathbb E|(H\mathbin\cdot M)_\infty|^2=\mathbb E\int_0^\infty H_s^2\,d[M]_s.}
$$
For every finite $t$, the same identity holds with the integral restricted to $[0,t]$ and the left side $\mathbb E|(H\mathbin\cdot M)_t|^2$. No claim of surjectivity onto all of $\mathcal M_0^2$ is needed: that would require an additional <Martingale representation theorem>.
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