Let $\mu, \nu$ both have finite second moments and $\nu \in \Pi(\mu, \nu)$. Then the inner product $\langle \cdot, \cdot\rangle$ is $\pi$-integrable

Question

In proving Knott-Smith optimality, I come across this result. Could you have a check if my attempt is fine, or it contains some logical mistakes?

Theorem: Let $\mu, \nu$ be Borel probability measures on $X:=\mathbb R^d$ such that $\mu, \nu$ both have finite second moments. Let $\pi$ be a Borel probability measure on $X^2$ with marginals $\mu, \nu$. Then the inner product $\langle \cdot, \cdot\rangle$ is $\pi$-integrable.

My attempt: Clearly, $\langle \cdot, \cdot\rangle$ is continuous and thus Borel measurable. Because $X^2$ is separable, $\langle \cdot, \cdot\rangle$ is $\pi$-measurable. We have $$ 2\langle x, y\rangle = |x|^2 + |y|^2 - |x-y|^2, $$ so $$ \begin{align} 2\int_{X^2} |\langle x, y\rangle| d \pi(x, y) &= \int_{X^2} \big | |x|^2 + |y|^2 - |x-y|^2 \big | d \pi(x, y) \\ & \le \int_{X^2} \big [ |x|^2 + |y|^2 + |x-y|^2 \big ] d \pi(x, y) \\ & \le 3 \int_{X^2} \big [ |x|^2 + |y|^2 \big ] d \pi(x, y) \\ & = 3 \int_{X} |x|^2 d \mu(x) + 2\int_{X} |y|^2 d \nu(y) \\ &< +\infty. \end{align} $$

This completes the proof.