A regular octahedron has edge length $2$. A plane through its center, perpendicular to the line joining the midpoints of two opposite edges, cuts the octahedron. What is the area of the cross-section?

1. Place the octahedron with vertices $(\pm\sqrt2,0,0),(0,\pm\sqrt2,0),(0,0,\pm\sqrt2)$ (edge $2$). The opposite edges from $(\sqrt2,0,0)$ to $(0,\sqrt2,0)$ and from $(-\sqrt2,0,0)$ to $(0,-\sqrt2,0)$ have midpoints on the line in direction $(1,1,0)$, so the plane is $x+y=0$.
2. The octahedron is $|x|+|y|+|z|\le\sqrt2$. The section is a rhombus with vertices $\left(\pm\frac{\sqrt2}2,\mp\frac{\sqrt2}2,0\right)$ and $(0,0,\pm\sqrt2)$.
3. Its diagonals are $2$ and $2\sqrt2$, so the area is $\frac12\cdot2\cdot2\sqrt2=\boxed{2\sqrt2}$.