Fillings of dispersion bands (E-K diagram) I struggle in understanding why in some references the bands filling by electrons in the E-k diagram is shown as an area delimited below by the dispersion curve and above by the Fermi energy (if in the last occupied band).
An example can be in Structure and Bonding in Crystalline Materials By Gregory S. Rohrer, p.356 (I linked the page on Google books since I do not know if there are some copyright issues by copying the figure).
I ever supposed the states were placed just over on the line (1D) (thus not having for the same energy level different k values) and that they were at a constant $\Delta k$ intervals.
Where am I wrong?
 A: I will try to answer in a simple and intuitive way. If you need more details I suggest you, for example, Ashcroft and Mermin, Solid State Physics.
The bands in the $E$-$k$ diagram are filled with electrons from the lowest energy state (the bottom of the dispersion curve) up to the Fermi energy.
This is a consequence of the Pauli exclusion principle and of energy minimization, and it is similar to the Aufbau principle of filling the electronic energy states in a single atom.
In fact, if your system has a number $N$ of electrons for a given unit cell of the crystal, these $N$ states will fill the bands beginning from the lowest energy level (energy minimization). For the Pauli principle, two electron cannot occupy the same quantum state (which is determined by the energy, the momentum $k$ and the spin), and therefore the occupied levels will "stack" one above the other in the $E$-$k$ level, up to the Fermi energy.
Indeed, the Fermi energy is defined by this construction, i.e., by the number of electrons in one unit cell $N$ and by the $E$-$k$ dispersion bands.
Anyway, electrons near the Fermi level are responsible for most of the electronic properties of the system, in particular current conduction. This is because electrons far below the Fermi level are in a way "frozen" in their energy level by the Pauli principle, since the only physical process they can undergo is the scattering (a jump in energy) to an energy level above the Fermi level $E_F$, where energy levels are unoccupied. And this costs an energy $\Delta E=E_F-E_i$, where $E_i$ is the initial energy of the electron deep under the Fermi level. As you can see, if $\Delta E$ is big (respect to thermodynamic fluctuations $k_B T$) this scattering process is nearly impossible (has a very small probability). 
If $\Delta E$ is small (respect to $k_B T$), that means that the electron level is near the Fermi level, and consequently this scattering is possible. The electron is not "frozen" in its energy state and, for example, can conduct current.
The $\Delta k$ you mention is related to the dimension of the system, and it is $\Delta k=2\pi/L$, where $L$ is the number of lattice sites of the system. If the system is large, $L\rightarrow \infty$ and $\Delta k\rightarrow 0$, and therefore the energy levels can be regarded as continuous lines (bands) as a function pf $k$, rather than a collection of isolated points.
Edit:
It is important to have in mind that in a 3-dimensional material (bulk metals, semiconductors, insulators), the Brillouin zone is 3-dimensional, and as a consequence the momentum is a vector $\mathbf{k}=(k_x,k_y,k_z)$. Hence the bands are a scalar function in a 3-dimensional domain, which describe the energy accessible to electronic states as a function of the momentum $mathbf{k}$, in the form $$E=E_b(\mathbf{k}),$$
where $b$ is the band index.
At a fixed energy $E$, the electronic states accessible describe a surface in the Brillouin zone. If $E=E_F$ (Fermi energy), this surface is the Fermi surface.
