Show this double integral is equal to $2\,(||f||^2_2||g||^2_2-|\langle f,g\rangle|^2)$

You just need to keep going.

\begin{align} \int_{I}\int_{I}|f(x)g(y) − f(y)g(x)|^2 \,\mathrm d x \,\mathrm d y&=\int_{I}\int_{I}(|f(x)g(y)|^2+|f(y)g(x)|^2-f(y)g(x)\overline{f(x)g(y)}\\&\qquad-f(x)g(y)\overline{f(y)g(x)})\,\mathrm d x \,\mathrm d y\\[0.3cm] &=\int_{I}\int_{I}(|f(x)|^2\,|g(y)|^2+|f(y)|^2\,|g(x)|^2-f(y)g(x)\overline{f(x)g(y)}\\&\qquad-f(x)g(y)\overline{f(y)g(x)})\,\mathrm d x \,\mathrm d y\\[0.3cm] &=\int_I|g(y)^2\,\bigg(\int_I|f(x)|^2\,dx\bigg)\,dy +\int_I|g(x)^2\,\bigg(\int_I|f(y)|^2\,dy\bigg)\,dx\\ &\qquad-2\int_I\int_If(y)\overline{g(y)}\,dy\,\overline{f(x)}\,g(x)\,dx\\[0.3cm] &=\|f\|^2\,\int_I|g(y)|^2\,dy+\|f\|^2\,\int_I|g(x)|^2\,dx-2\langle f,g\rangle\,\int_I\overline{f(x)}g(x)\,dx\\[0.3cm] &=\|f\|^2\|g\|^2+\|f\|^2\|g\|^2-2\langle f,g\rangle\,\overline{\langle f,g\rangle}\\[0.3cm] &=2\|f\|^2\|g\|^2-2|\langle f,g\rangle|^2. \end{align}