How to determine the degeneracy of an energy level for a periodic quantum system from its band structure over the Brillouin zone? 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?

 A: You are definitely on the right track, but the accounting here is a little tricky and easy to mistake.
First of all, and probably most important, there is no difference between the valence and conduction bands at the K points. They are the same; they don't count twice. Of course, like any state, two electrons can fill it with opposite spins.
So then, why are there two K points (and not one or six)?  Try the trick for counting atoms in a unit cell, when there are atoms at the edges of the cell (are they in or out? do they count or not?): Imagine shifting the cell boundaries slightly in any direction, keeping the atoms fixed. Then, how many atoms are contained wholly within the cell volume? 
The trick works in reciprocal space as well. Do the same for graphene's Brillouin zone, count the K points, and you'll find two, next to each other. By symmetry, these are the same as the ones opposite each other. Alternatively, if you think about it, each K point is really only $1/3$ within the Brillouin zone, and $1/3*6=2$.
So for counting the states for the density of states, for example, you'll find a total degeneracy factor of 4: two for the K points, and two for the spins. 
