If you want to go further in understanding this point of view, I would advise you to begin learning class field theory. It is a deep subject, it can be understood in a vast variety of ways, from the very concrete and elementary to the very abstract, and although superficially it appears to be limited to describing abelian reciprocity laws, it in fact plays a crucial role in the study of non-abelian reciprocity laws as well.
The texts:
Ireland and Rosen for basic algebraic number theory, a Galois-theoretic proof
of quadratic reciprocity, and other assorted attractions.
Cox's book on primes of the form x^2 + n y^2 for an indication of what some of the content of class field theory is in elementary terms, via many wonderful examples.
Serre's Local Fields for learning the Galois theory of local fields
Cassels and Frolich for learning global class field theory
The standard book at the graduate level to learn the arithmetic of non-abelian (at least 2-dimensional) reciprocity laws is Modular forms and Fermat's Last Theorem, a textbook on the proof by Taylor and Wiles of FLT. But it is at a higher level again.
I don't think that you will find a single text on this topic at a basic level (if basic
means Course in arithmetic or Ireland and Rosen), because there is not much to say beyond what you stated in your question without getting into the theory of elliptic curves and/or the theory of modular forms and/or a serious discussion of class field theory.
Also, as basic suggests, you could talk to the grad students in your town, if not at your institution, then at the other one down the Charles river, which as you probably know is currently the world centre for research on non-abelian reciprocity laws (maybe shared with Paris). Certainly there are grad courses offered on this topic there on a regular basis.
The proof given by Steinhaus (following A. Schinzel) is slick indeed. I'll paraphrase what he does:
At step number $n$, the interval (0,1) is divided into $n$ intervals $I_0^{(n)}=(0,\frac{1}{n}), \ldots, I_{n-1}^{(n)}=(\frac{n-1}{n},1)$. Then $\lfloor nP \rfloor$ gives you the number of the interval to which a point $P$ in the sequence belongs in the $n.$ step.
He proceeds by proving the following claim:
If $P_1 \in (\frac{7}{35},\frac{8}{35})$ and $P_2 \in (\frac{9}{35}, \frac{10}{35})$, then there is some $n \in \mathbb{N}$ big enough such that $\lfloor (n-1)P_1 \rfloor < \lfloor nP_1 \rfloor$ , but $\lfloor (n-1)P_2 \rfloor = \lfloor nP_2 \rfloor$.
Let's see how this claim implies that there is no infinite such sequence. Eventually, in your sequence at the $k.$ step there will be two points $P_1,P_2$ as in the claim. Let $n$ be as in the claim, in particular $n>k$.
Let $k_1:=\lfloor (n-1)P_1\rfloor$ and $k_2:=\lfloor (n-1)P_2\rfloor$. Then $P_i$ belongs to the $k_i$. interval in the $(n-1).$st step and there are $k_2-k_1-1$ points of $a_1, \ldots, a_{n-1}$ of the sequence in-between. But in the $n.$ step, $P_1$ will move to the $k_1+1$.st interval, while $P_2$ will remain in the $k_2$.st interval, and so there will be a ''collision'' of the in-between points.
Best Answer
Nope! Amazingly enough, no elementary proof of this fact is yet known (Edit: See KConrad's answer). The difficulty is tied up in some pretty fantastic algebraic/analytic number theory, namely the analytic class number formula. But, without getting into that, here's the short form of the story: let $L(s)$ be the $L$-function attached to the character arising from the Kronecker symbol mod $q$. Then one can compute analytically via the Euler product formula for this $L$-function that (for an explicit positive constant $C$), $$ L(1)=C\left[\sum_{m=0}^{q/2}\left(\frac{m}{q}\right)-\sum_{m=q/2}^{q}\left(\frac{m}{q}\right)\right]=\frac{\pi}{\left(2-\left(\frac{2}{q}\right)\right)q^{1/2}}\sum_{m=0}^{q/2}\left(\frac{m}{q}\right), $$ the positivity of which gives the desired statement about the distribution of quadratic residues.
For a reference (from whence I pulled this out of), Davenport's "Multiplicative Number Theory" is pretty fantastic. This is all done in the first 4-5 pages.