On the quadratic variation of a martingale

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I am currently trying to understand a proof about the construction of the Ito integral, and there is a step I am having serious trouble with. Here, $(\mathcal{F}_t)$ is a filtration, $M$ is a square integrable martingale starting at $M_0=0$, and $X$ is an adapted, cadlag, bounded process. Of course, since $M$ is a square integrable martingale, we have $$\mathbb{E} M_t^2 = \mathbb{E} [M]_t,$$ where $[M]_t$ is the quadratic variation of $M$. The author now uses that $$\mathbb{E}\big( X_s^2(M_t-M_s)^2 \big) = \mathbb{E}\big( X_s^2([M]_t-[M]_s)\big).$$ Without the $X_s^2$, this is clear to me, but I do not see why this holds here. I have tried conditioning on $\mathcal{F}_s$, but that also did not help, since $M$ is a general martingale.

I would be very thankful if somebody could clear this up for me.