(1) For the existence of $\mathfrak{p}$, here's what I'm doing. Given any nonzero number $a$, there exist only finitely many prime ideals containing $a$. So given a finite number of $a_i$, I can find a prime not containing any of them. (2) I guess the spirit is similar to Euclid's proof, but it's slightly different. I already know that there are infinitely many prime ideals, and I'm using it to run the argument.

Do you mean, there are infinitely many irreducible elements that don't differ by a unit? If so, why can't we pick the irreducible inductively? If we have $a_1,\dotsc,a_k$, then pick a prime $\mathfrak{p}$ not containing any of the $a_i$, pick an element $b \in \mathfrak{p}$, and decompose until we get $a_{k+1} \in \mathfrak{p}$ irreducible.

If we fix $a, q$ and increase $x$, isn't this essentially the Gauss circle problem with some shift in the lattice?

When you say that 2-dimensional slices are of rank 1, do you mean that its image under $\mathrm{id} \otimes \mathrm{id} \otimes \xi$ is of rank 1 for all linear functionals $\xi : V \to k$, or for only linear functionals $\xi$ that are projection to the axes?

I'm a bit confused; why isn't this true just by definition after expanding out?