On Bloch's Theorem and the standard Fourier proof I am reading Ashcroft and Mermin and I am unfortunately finding the (standard) proof of Bloch's theorem presented there rather befuddling. The proof in its entirety is at the bottom of this post. I have included it not to be lazy, but rather to provide a prospective answerer with the context of what I am confused about.
I want to also note that I have subsequently consulted textbooks by Simon, Marder, and Ibach and Luth, but to no avail. Also, I have found some Physics SE answers informative (in particular, the first answer given in Deriving Bloch's Theorem), but none completely satisfying. It is for that reason that I am asking this question.
Now, to present what I've done in analyzing the given derivation of Bloch's theorem:
To situate ourselves, we note that our general program will be to produce a (complete) basis of the Hilbert space of one electron in a potential with the periodicity of some given Bravais lattice. Now I have no problems with the steps up to and including Equation (8.37). I would though like to make clear my thinking: we are writing down (8.30) as the general form for an eigenstate (indeed, any state) of the system's Hamiltonian, and will plug it in ansatz to the Hamiltonian eigenvalue equation to get constraints on its form.
My first question arises in the move to (8.38). In Deriving Bloch's Theorem, the first answer notes that we ought to distinguish between the cases in which $\{\mathbf{q}\}$ and $\{\mathbf{q+K}\}$ coincide, and when they do not. I agree with that user's conclusions in both cases, but am not clear on why these are the only two cases in general. Why can there not exist only some $\mathbf{q}$ such that $\mathbf{q}=\mathbf{q'+K}$ for some $\mathbf{q'}$?
At any rate, that is not the main concern for me. My main confusion surrounds the steps beginning "Putting this information back into the expansion (8.30)..." on page 139. A&M seem to be saying that if the eigenstate is such that it can be built up of $\mathbf{k}$, $\mathbf{k+K}$, etc., then the eigenstate has the desired Bloch form. But this seems to be begging the question?



 A: When a function is periodic along a lattice $\mathcal L$, then it can be expanded as a Fourier series over the reciprocal lattice $\tilde{\mathcal L}$.
The potential $U$ is periodic of the the Bravais lattice (let's take a basis $a_i$), so its Fourier series sums over $\mathbf K$, which belongs to the reciprocal Bravais lattice (which is generated by the reciprocal basis $\mathbf b_i$).
Meanwhile, the Born-Von Karman boundary conditions means that we take $\psi$ to be periodic over a lattice generated by $N_i \mathbf a_i$ for some (large) integers $N_i$. Its Fourier series then sums over $\mathbf q$, which belongs to the lattice generated by the vectors $\mathbf b_i/N_i$. This clearly includes the reciprocal Bravais lattice, and therefore, for any fixed $\mathbf K'$, $\mathbf q\mapsto \mathbf q - \mathbf K$ is a perfectly valid change of index.
In other words, because of the Born-Von Karman boundary conditions, we always have $\{\mathbf q\} = \{\mathbf q + \mathbf K\}$.
Also, the mapping $(\mathbf k,\mathbf K)\mapsto \mathbf q= \mathbf k - \mathbf K$, with $\mathbf k$ in the first Brillouin zone and $\mathbf K$ in the reciprocal Bravais lattice, is one-to-one and onto : we can always decompose $\mathbf q$ as $\mathbf k-\mathbf K$ in a unique way.
Then at that point, A&M notice that the equations contraining the coefficients $c_{\mathbf q}$ split into independent equations contraining the $c_{\mathbf k -\mathbf K}$ with $\mathbf k$ fixed in the first Brillouin zone. Therefore, the general solution to $(8.41)$ is a superposition of the $\psi_{\mathbf k}$ given in $(8.42)$.

Edit
When studying the Born-Von Karman boundary conditions, A&M use Bloch's theorem to show that the periodicity lattice imposed on the system must be generated by vectors of the form $N_i \mathbf a_i$. This is actually a bit overkill : if you want the translation operators $\hat T_{\mathbf a_i}$ to be  well defined with the periodic boundary conditions, you see that this condition need to hold, without appealing to Bloch's theorem.
When doing the change of variable $\mathbf q\to (\mathbf k , \mathbf K)$, the set of values for $\mathbf k$ need to contain exactly one vector of the for $\mathbf q - \mathbf K$ for any fixed $\mathbf q$ (ie one member of each equivalence class modulo the reciprocal Bravais lattice). The first Brillouin zone is precisely defined to be such a set.
We are trying to solve the time-independent Schrödinger equation. We know that the solutions can be chosen to be an orthonormal basis. When we reduce the TISE to $(8.41)$, we see that any solution can be written as a linear combination of solutions of the form $(8.42)$, i.e. we can look for solutions with a well-defined quasi-momentum $\mathbf k$.
Choose one $\mathbf k$ in the first Brillouin zone and solve $(8.42)$ for $\psi_{\mathbf k}$ given by $(8.42)$. You will find several solutions, which we label $\psi_{n\mathbf k}$ and with energies $E_n(\mathbf k)$. Since the subspace of states of the form $(8.42)$ is still of infinite dimension, we expect to find a countable infinity of solutions $\psi_{n\mathbf k}$.

Edit 2
About that last point. For a fixed $\mathbf k$, wave-functions of the form $(8.42)$ are the ones with $\psi(\mathbf x + \mathbf R) = e^{i\mathbf k \cdot \mathbf R}\psi(\mathbf x)$ (for any $\mathbf R$ in the Bravais) along with the Born-Van Karman boundary condition. This space is infinite dimensional and we are looking for an orthonormal basis of eigenstates : we therefore expect to find infinitely many.
For example, this is the case for nearly free electrons or in the tight-binding regime.
