= Solution
For this <Cartesian product>, the requested $U^2$ norm is the <box norm>, with complex conjugates:
$$
\boxed{\|f\|_{U^2}^4=\mathbb E_{x,x'\in X}\mathbb E_{y,y'\in Y}
f(x,y)\overline{f(x',y)}\overline{f(x,y')}f(x',y').}
$$
Equivalently it is $\mathbb E_{y,y'}|K(y,y')|^2$, where $K(y,y')=\mathbb E_xf(x,y)\overline{f(x,y')}$. This makes its nonnegativity explicit.
Apply the <Cauchy-Schwarz inequality> first in $x$:
$$
|\mathbb E_{x,y}f(x,y)u(x)v(y)|^2
\le\|u\|_2^2\,\mathbb E_{y,y'}v(y)\overline{v(y')}K(y,y').
$$
The right-hand expectation is nonnegative, since it came from $\mathbb E_x|\mathbb E_yf(x,y)v(y)|^2$. A second application in $(y,y')$ gives
$$
|\mathbb E_{x,y}fuv|^4
\le\|u\|_2^4\,
\mathbb E_{y,y'}|v(y)\overline{v(y')}|^2
\mathbb E_{y,y'}|K(y,y')|^2
=\|u\|_2^4\|v\|_2^4\|f\|_{U^2}^4.
$$
Taking fourth roots proves the <bilinear correlation bound for the box norm>
$$
\boxed{|\mathbb E_{x,y}f(x,y)u(x)v(y)|\le\|f\|_{U^2}\|u\|_2\|v\|_2.}
$$
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