So I have found this link which I will try after writing this post, but I would like to see if my original attempt (which is his/her attempt there) can be made to work. The reason I want this to work is that this works for $M = $ Brownian motion.
To prove: for $M$ a $L^2$-bounded continuous martingale, $$\newcommand{\E}{\mathbb{E}} \E\left[\int_0^∞K_s \ dM_s\right]^2 \overset{?}{=} \E ∫_0^∞ K_s^2 \ d⟨ M,M ⟩ _s $$ which for the case of $K$ elementary(called 'simple' in that link) with $K_i ∈ \mathcal{F}_{t_i}$, reduces to $$\E\left [\sum_i K_i (M_{t_{i+1}} - M_{t_i})\right]^2 \overset{?}{=} \E \sum_i K_i^2 (⟨ M,M ⟩_{t_{i+1}} - ⟨ M,M ⟩_{t_i}) $$
My strategy is to expand the sum on the left into $$\sum_{i≠ j}K_iK_j(M_{t_{i+1}} - M_{t_i})(M_{t_{j+1}} - M_{t_j}) + \sum_{i}K_i^2(M_{t_{i+1}} - M_{t_i})^2$$ The second sum is going to be what we want, so we hope the first sum on expectation is 0,
\begin{align}&\E\sum_{i≠ j}K_iK_j(M_{t_{i+1}} - M_{t_i})(M_{t_{j+1}} - M_{t_j}) \\ =& 2\sum_{i<j} \E K_iK_j(M_{t_{i+1}} - M_{t_i})(M_{t_{j+1}} - M_{t_j}) \\ =& 2\sum_{i<j} \E K_i (M_{t_{i+1}} - M_{t_i})\color{red}{\E[ K_j(M_{t_{j+1}} - M_{t_j}) | \mathcal{F}_{t_i}] }\end{align}
At this point, were $M$ Brownian, we would be able to apply the fact that restarted Brownian motion is independent of the $\sigma$-algebra of the time you restarted, i.e. $$ M_{t_j+s} - M_{t_j} \perp \!\! \perp \mathcal{F}_{t_j} $$ so we have the splitting $$ \E[ K_j (M_{t_j+s} - M_{t_j})|\mathcal{F}_{t_i}] =\E[ K_j] \E[ (M_{t_j+s} - M_{t_j})|\mathcal{F}_{t_i}] = 0 $$
Which proves the result. Can this be fixed for arbitrary continuous martingales?
