The following figure shows the 1st Brillouin zone of graphene (shaded area).
At the $K$ and $K'$ points (called Dirac points), the upper (conduction) band touches the lower (valence) band, and therefore at the Dirac points, we have a two-fold degeneracy (valence band and the conduction band are degenerate with zero energy). We know that all three $\mathbf K$ points are equivalent, and also all $\mathbf K'$ points are equivalent. Therefore, we only have two distinct Dirac points, each of them being degenerate with $E=0$.
Now, these two Dirac points, which are located at $\mathbf K = -\mathbf K'$ (as a result of TR-symmetry) both have the same energy (zero) and are degenerate too. Therefore, there are 4 states that have zero energy (conduction and valence bands, at K and K' points), and the zero energy state should be 4-fold degenerate.
But in various resources, I've read that in graphene "there is one pair of Dirac points, and the zero-energy states are doubly degenerate; which is also called a twofold valley degeneracy".
What was wrong with my reasoning? Why are zero-point energy states of graphene two-fold degenerate (instead of four-fold)? In particular, aren't all points of the Brillouin zone with the same energy (and of all bands) counted toward the degeneracy of an energy level?