Solution:

Let the cube have side length $a$. The area of a face is $a^2$.

Any cross-section through the center of the cube is a plane passing through the center. The largest possible cross-section is when the plane is perpendicular to the space diagonal, which gives a regular hexagon. The smallest possible cross-section is when the plane is parallel to a face, which gives a square of area $a^2$.

Let us show that any cross-section through the center has area at least $a^2$.

Let the cube be centered at the origin, with faces parallel to the coordinate planes. Consider a plane passing through the origin with normal vector $(l, m, n)$, where $l^2 + m^2 + n^2 = 1$.

The intersection of this plane with the cube is a convex polygon. The area of the cross-section is maximized when the plane is perpendicular to the space diagonal, and minimized when the plane is parallel to a face.

Let us compute the area:

The area of the cross-section is $A = \frac{a^2}{|n|}$, where $n$ is the direction cosine of the normal with respect to the axis perpendicular to the face. Since $|n| \leq 1$, $A \geq a^2$.

Alternatively, by symmetry, the minimal area occurs when the plane is parallel to a face, and the cross-section is a square of area $a^2$.

Therefore, any cross-section through the center has area not less than the area of a face.