Well Zurek's proof is very simple.
The idea is take a Schmidt state:
$$\psi = \Sigma_{i} \alpha_{i} |\sigma_{i}\rangle |\eta_{i}\rangle \tag 1$$
with ##|\sigma_{i}\rangle## being states of the microscopic system and ##|\eta_{i}\rangle## being environment states.
The state is said to be envariant under a transformation, ##U_{s}##, of the microscopic system, if it can be undone by a transformation, ##U_{\eta}##, of the environment .
Lemma 1: For states with ##|\alpha_i| = |\alpha_k|## for some ##i,k##, then swapping ##|\sigma_{i}\rangle## and ##|\sigma_{k}\rangle## is an envariant transformation.
That is, for Schmidt states, swaps of microsystem states with equal magnitude coefficients can be undone by the environment.
Next Zurek has four axioms, with a theorem that can be proved from the first three.
A1. To represent an alteration of a system ##S##, a unitary transformation must act on its Hilbert Space ##\mathcal{H}_{S}##
A2. All information on observables, their values and their probabilities for a system ##S## is fully captured by the state of ##S##. This state might be mixed or pure.
A3. The state of a subsystem is fully specified by the state of the total system.
From these three axioms you can conclude:
Theorem 1: Phases do not affect probabilities, i.e. probabilities depend only on ##|\alpha_{i}|##.
The final axiom comes in three versions, anyone of them may be added to the list as the final axiom.
B1. If swapping two orthogonal states leaves the state of the system ##S## unchanged, the probabilities of the outcomes associated with those states are the same.
B2. If all unitary transformations within a subsystem ##\bar{S}## of ##S## leave ##S## unchanged, then the probabilities of any state in an orthonormal basis of ##\mathcal{H}_{\bar{S}}## are equal.
B3. The probabilistic meaning of a Schmidt state is that the environment and the state are perfectly correlated, i.e. observing ##|\sigma_{i}\rangle## means ##|\eta_{i}\rangle## will be observed with probability 1.
Any one of these axioms allows him to prove the following:
Theorem 2: Terms with equal amplitudes in Schmidt states like ##(1)## have equal probabilities.
Corollary: In a Schmidt state with all ##N## coefficients equal, all outcomes are equally likely, i.e. ##1/N##.
Looking ahead I will say that using B2 permits you to prove this without using A2.
From there he uses the fact that the environment can always be enlarged by adding an extra system to reduce the unequal amplitude case to the equal amplitude case. As an example say we have:
$$\psi = \sqrt{\frac{1}{3}}|\sigma_1\rangle|\eta_1\rangle + \sqrt{\frac{2}{3}}|\sigma_2\rangle|\eta_1\rangle \tag 2$$
We can introduce another system, ##\beta##, essentially enlarge the environment so that this becomes:
$$\psi = \sqrt{\frac{1}{3}}|\sigma_1\rangle|\eta_1\rangle|\beta_1\rangle + \sqrt{\frac{2}{3}}|\sigma_2\rangle|\eta_2\rangle|\beta_2\rangle \tag 3$$
Then provided the new environment ##\beta## is large enough so that most of its eigenstates (the states that couple to the microscopic system) are degenerate, we can expand them enough to counterweight the unequal amplitudes:
$$\beta_1 = \gamma_1 \tag 4$$
$$\beta_2 = \sqrt{\frac{1}{2}}\gamma_{2,1} + \sqrt{\frac{1}{2}}\gamma_{2,2} \tag 5$$
and so, substituting ##(4),(5)## into ##(3)##:
$$\psi = \sqrt{\frac{1}{3}}|\sigma_1\rangle|\eta_1\rangle|\gamma_1\rangle + \sqrt{\frac{1}{3}}|\sigma_2\rangle|\eta_2\rangle|\gamma_{2,1}\rangle + \sqrt{\frac{1}{3}}|\sigma_2\rangle|\eta_2\rangle|\gamma_{2,2}\rangle$$
Hence this is now an equal amplitude case, and by Theorem 2 all have equal probability ##1/3##. Since ##\sigma_2## appears twice, we can say it has ##2/3## chance of being seen.
And so Zurek obtains the Born rule for amplitudes that are roots of rationals.
To obtain it for all reals, he uses the fact that ##\mathbb{Q}## is dense in ##\mathbb{R}## and that an "arbitrarily" fine grained* larger environment can be found.
Thus we have the Born rule. Issues to follow.
*In the sense of having as large as necessary expansion in the form of ##(5)##
if there are such things. We shall see, And I'll let someone else digest the proffered argument for me!