Why the electrons below the Fermi level do not conduct electricity?
Electrons in Solids
Electrons in solids are treated by the so-called Band Theory, a quantum mechanical model for fermions interacting with a fixed crystal lattice. The solution to the Schrodinger's equation shows that atomic orbitals overlaps and the energy levels of the electrons form energy bands of a few electron volts each. If the lattice has $N$ atoms, then each band has $N$ k-states (momentum states) which are distributed symmetrically between positive and negative momenta. Since the number of atoms in solids is typically large, the energy levels in each band is so close that it looks like a continuum. The bands are separated by a finite energy, i.e. regions empty of states called bandgaps (or simply gaps). It is the filling of these bands and the size of the gaps that characterize electrical conductivity (as well as optical properties) of solids.
Let us assume that we are at zero (absolute) temperature. The electrons fill the bands in a way that energy is minimized but since they obey Pauli Exclusion Principle, there can be only two electrons (spin up and spin down) in each k-state. Thus the electrons occupy the bands from lowest to highest energy levels. At $0\, \mathrm K$ all the levels are completely filled from the bottom. The last band with electrons is called valence band and the energy of the highest occupied level is then said to be the Fermi energy.
Electrical Conductivity from Band Theory
Note that all the electrons below the Fermi level cannot decay, all the states bellow are occupied and the electrons must satisfy the Pauli Principle. By the same reason, the only way they can be excited is by going above the Fermi level, if that is possible. Sometimes people refer to these electrons as being frozen.
When an electric field is applied to the solid, the electrons interact with the field and therefore can change their state. However an electron can only gain energy from the field if there is an available state for it to jumping on. Suppose we start with an arbitrarily weak field, the electrons deep bellow the Fermi level cannot be excited, the energy possibly gained from the field is not enough to bring it to an empty state. On the other hand, the electron near the Fermi level may or may not be excited and that is what distinguish between conductors and insulators.
Let us say that our solid has one valence electron, i.e. the $N$ atoms have one electron in the last shell each. As we saw, each band has $2N$ states so that the valence band in only half filled. By the symmetric distribution of the k-states, the average momentum of the electrons is zero. This means they are not moving in a preferred direction, there is no current flowing.
If an electrical field is applied along some direction, say the $x$ axis, the electrons can accelerate since there are plenty of empty states to go. There will be a repopulation of electronic states. This means that electrons with negative k-states migrate to positive ones, which are above the Fermi level. Note that the average momentum is now non zero, there is a preferred direction to move which gives raise to a non zero current. This solid is a conductor.
On the other hand, if the valence band is completely filled, there can be no repopulation because there is a gap above the Fermi level. The average momentum is still zero and this solid does not conduct.
As we can see from this two examples above, when we say that only electrons above the Fermi level conduct what we really mean is that in order to have electrical conductivity there must be a repopulation of k-states and this is only possible if the electrons can jump to available states, the states above the Fermi level.
The Pauli exclusion principle is the key thing here. Focusing on momentum space and looking at isotropic space (not possible in a crystal, of course) at first. At cold temperatures, most electrons will try to get into the state of minimal energy. There is only one state with zero momentum. The next electron has to choose a state with non-zero momentum. The third electron will then choose the state on the other side of the momentum state such that their momenta add up to zero. Adding more and more electrons will result in a Fermi sphere, look at the image on this page for a representation. It is just a sphere in momentum space.
There are electrons moving the whole time, but always the same number into the opposite direction as well. So although they move, there is no net current. In a crystal, the Fermi surface will not look that simple, but more like the one shown on that page for copper. The Fermi surface cannot be compressed due to the Pauli exclusion principle.
When you apply a tiny external electric field, the states moving along the electric field will be favored. When all electrons start to drift a bit, the whole Fermi sphere in momentum space will be shifted to the side. Then it becomes asymmetric (in momentum space) and there will be a net current.
Looking at the differences between the Fermi surfaces with and without the external field, you will see that the bulk has not changed (electrons are indistinguishable) but that the surface has moved. One one side they are above the original Fermi surface, on the other the occupied states might no longer reach the Fermi surface.
At $T = 0$, all states below the Fermi surface are occupied, none with higher energy are occupied. With $T > 0$, the distribution is as $\exp(- E/T)$. Higher states can now be occupied while there are free states at lower momentum within the sphere. Therefore the Fermi surface will start to blur out.