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.