Why aren't all bonds in cyclobutadiene not equal?
Strictly speaking cyclobutadiene is not antiaromatic but non aromatic, ie the same energy 4A + 4B is attained by applying the Hückel LCAO, as for two isolated double bonds, 2 times 2A + 2B.

Given that cyclobutadiene can only be isolated as non delocalized system when in matrixes and a very low T points to strain as a reason for instability of both forms, the delocalized one remaining hypothetical.

Edit for the readers not familiar with the terms: A system obeying all the requisites (cyclic, pi orbitals on each atoms) is antiaromatic when the hypothetical delocalization destabilizes it as compared to the localized system. If no energy difference arises, the system is called non aromatic.

Beware there are plenty of pages discussing non aromaticity in a trivial way, e.g. aliphatic or linear compounds. Cyclobutadiene is non aromatic within the Hückel picture. Also it seems worth to point out again that antiaromatic refers anyway to an hypothetical structure, not to an actual delocalization. A system which cannot avoid it by distorting would be unstable tout-court. Cyclobutadiene, albeit being non aromatic, is on the same line. Indeed you can find antithetical assertions about it, such as it being "the archetype of antiaromatics“ along with “it reacts as diradical“ as if the two things would be straight forward correlated. Normally a system escape antiaromaticity. Being cyclobutadiene non aromatic results in a distorted rectangular molecule, but only when it cannot escape this anyway unstable situation. See also Resonance structure of cyclobutadiene?. Although the situation of cyclobutadiene is again erroneously described as an antiaromatic case, it gives the answer. Also worth reading is the answer by Georg.