 Solution

Let $X$ be a subcomplex of a product of graphs $A\times B$. Orient every edge of each factor. An edge of $X$ is horizontal or vertical according to its factor, and this type is preserved across opposite sides of every square.

A hyperplane of a cube complex of horizontal type retains one fixed edge of $A$ while moving through edges of $B$; the analogous statement holds vertically. The factor orientation makes every hyperplane two-sided. A square has one horizontal and one vertical direction, so no hyperplane self-intersects. At a vertex there is at most one incident edge with a fixed factor edge and orientation, so no hyperplane self-osculates. Finally, a horizontal hyperplane and a vertical hyperplane can cross only in the unique product square determined by their two factor edges. If that square belongs to $X$, it fills every corner at which those two dual edges meet; if it does not, the hyperplanes never cross. Thus no pair interosculates. All four hyperplane pathologies are absent, so $X$ is a special cube complex.