More generally, let $1 \leq a \leq b \leq n$ and $\mathcal{F}_k = \sigma (X_1, \cdots, X_k)$ be the filtration generated by $(X_k)$. If $Y$ is $\mathcal{F}_a$-measurable and $(X_b - X_a)Y$ is integrable, then we have $ \mathbb{E}[ (X_b - X_a) Y ] = 0. $ Indeed, by tower property of the conditional expectation, $ \mathbb{E}[ (X_b - X_a) Y ] = \mathbb{E}[ \mathbb{E}[ (X_b - X_a) Y | \mathcal{F}_a ] ]. $ Since $Y$ is $\mathcal{F}_a$-measurable, we can take it out and thus $ \begin{align*} \mathbb{E}[ (X_b - X_a) Y ] & = \mathbb{E}[ \mathbb{E}[ (X_b - X_a) Y | \mathcal{F}_a ] ] \\ & = \mathbb{E}[ \mathbb{E}[ X_b - X_a | \mathcal{F}_a ] Y ] \\ & = \mathbb{E}[ (\mathbb{E}[ X_b | \mathcal{F}_a ] - X_a) Y ]. \end{align*}$
But now we know that $\mathbb{E}[ X_b | \mathcal{F}_a ] = X_a$ by tower property again and induction as follows: $ \mathbb{E}[ X_b | \mathcal{F}_a ] = \mathbb{E}[ \mathbb{E}[ X_b | \mathcal{F}_{b-1} ] | \mathcal{F}_a ] = \mathbb{E}[ X_{b-1} | \mathcal{F}_a ] = \cdots = \mathbb{E}[ X_{a+1} | \mathcal{F}_a ] = \mathbb{E}[ X_a | \mathcal{F}_a ] = X_a. $
This completes the proof of the claim.