The general argument so far is great in that it applies to virtually any complicated electron system with interactions and in a real material, but we would probably feel better if we could calculate the Hall conductance directly for some simple system. So let us try to do this for the simplest case of electrons in a magnetic field.
For starters, let us forget about the Corbino disk and just ask what do quantum mechanical electrons do in a magnetic field.
We know what classical electrons do in a perpendicular magnetic field: They go around in cyclotron orbits, because of the Lorentz force. The cyclotron radius in a magnetic field of strength \(B\) for an electron with velocity \(v\) is \(r_c = mv/eB\). An electron performing a cyclotron orbit at velocity \(v\) has angular momentum \(L=mvr_c=eB r^2_c\). In quantum mechanics, however, only orbits with a quantized angular momentum \(L=n\hbar\) will be allowed. From the equality \(r^2_c = n\hbar/eB\) one obtains that only some discrete values are allowed for the radius, \(r_n = \sqrt{n} l_B\), where \(l_B = \sqrt{\hbar/eB}\) is called the magnetic length.
All cyclotron orbits, independent of the radius, circle at the same frequency \(\omega_c=eB/m\). The energy of the electron in this quantized orbit is equal to \(L\omega_c = n\hbar\omega_c\). So the energy spectrum really looks like that of a harmonic oscillator. All the energy levels are also shifted up from zero energy by the zero-point motion of the harmonic oscillator, \(\hbar\omega_c/2\). We finally obtain that the allowed energy levels are
\[E_n = \hbar \omega_c \,\left(n+\tfrac{1}{2}\right)\,.\]
These quantized energy levels of electrons in a magnetic field are called Landau levels.
You can put many electrons in the same Landau level: one for every flux quantum of the magnetic flux passing through the system. Therefore Landau levels have a huge degeneracy, proportional to the area of the sample.
Now that we know the answer in advance, we can solve the Schrödinger equation for electrons in a magnetic field without stress. It will still be important to understand the quantum Hall effect in a bit more detail. The Hamiltonian is
\[H=(\textbf{p}-e \textbf{ A})^2.\]
The vector potential \(\bf{A}\) depends on position, which makes this Hamiltonian complicated to solve in general. For a uniform magnetic field, we can make our life easier by choosing a Landau gauge
\[\textbf{A}(x,y)=\hat{\textbf{x}}B y ,\]
where the vector potential does not depend on \(x\). In this gauge, the entire Hamiltonian is translationally invariant along the \(x\) direction, and therefore commutes with the corresponding momentum \(p_x\). This allows us to choose \(p_x=\hbar k\) as a good quantum number, and our two dimensional Hamiltonian reduces to a one dimensional one:
\[H(k)=p_y^2+(\hbar k-e B y)^2.\]
Apart from a shift of the \(y\) coordinate by \(y_0(k)=\hbar k/eB\), this is exactly the Hamiltonian of a simple harmonic oscillator! Its eigenvalues are the Landau levels, which are independent of \(k\). The corresponding wave functions are those of the harmonic oscillator along the \(y\) direction, and plane waves with momentum \(k\) along the \(x\) direction. In the \(y\) direction, they are localized in space within a length \(\sim l_B\).
This gives us another way to understand the quantized Hall conductance for ideal two dimensional electron gases.
Now, the electron energies are quantized in Landau levels, and if \(n\) Landau levels are filled at a given chemical potential, the filling factor is \(\nu=n\). The Streda formula then predicts the Hall conductance as \(\sigma_H=\nu e^2/h=n e^2/h\). The longitudinal conductivity must vanish since the gapped system does not allow dissipation of energy in the bulk.