 Solution

Call the two horizontal edge classes indicated by one and two arrowheads $a$ and $b$. All vertices in the displayed quotient are identified. In the third displayed square, an $a$-edge and a $b$-edge are opposite, so they are dual to the same hyperplane of a cube complex $H$. At the unique vertex, the distinct edges $a$ and $b$ are therefore dual to $H$, but no square has them as adjacent sides. With the orientations shown, they have the same initial vertex. Hence $H$ is a self-osculating hyperplane, one of the forbidden pathologies of a special cube complex. The displayed cube complex is not special.