Trouble understanding the proof that the orbit of every $x\in S^1$ is dense under the mapping $R_\alpha x = x + \alpha \mod 1$ for irrational $\alpha$ The following is taken from the page four of the book Introduction to Dynamical Systems by Brin and Stuck.
The authors are considering the unit circle $S^1 = [0, 1]\setminus \sim$ where $0$ and $1$ are identified with distance on $[0, 1]$  as $d(x, y) = \min\{|x - y|, 1 - |x - y|\}$. Let the mapping $R_\alpha$ be defined as $R_\alpha x = x + \alpha \mod 1$ for $x \in S^1$. Below is the proof the authors give on why the orbit of every $x \in S^1$ is dense if $\alpha$ is irrational:

[...] if $\alpha$ is irrational, then every positive semiorbit is dense in $S^1$. Indeed, the pigeon-hole principle implies that, for every $\epsilon > 0$, there are $m, n < \frac{1}{\epsilon}$ such that $m < n$ and $d(R^m_\alpha, R^n_\alpha) < \epsilon$. Thus $R^{n - m}$ is rotation by an angle less than $\epsilon$, so every positive semiorbit is $\epsilon$-dense in $S^1$ (i.e. comes within distance $\epsilon$ of every point in $S^1$).

I am assuming that when the authors write

$d(R^m_\alpha, R^n_\alpha)$

they mean

$d(R^m_\alpha(x), R^n_\alpha(x))$

What I don't understand is
1.) What precisely is the used pigeonhole argument? That given $\epsilon > 0$, we may divide the circumference of $S^1$ into disjoint $\lfloor \frac{2\pi}{\epsilon}\rfloor$ intervals so that given a point $x \in S^1$, by ??? two orbits originating from $x$ must lie in the same interval?
2.) How exactly is the density of an arbitrary orbit shown? So far my work with dense sets has been limited to the definition:

$A$ is said to be a dense subset of $B$ in topology $\tau$ if for every open subset $U$ of $B$ we have $A\cap U \neq \varnothing$.

 A: To simplify things let us assume that our circle has circumference equal to $1$ and each time you apply $R_\alpha$ the point shifts by arc length $\alpha$ in anti-clockwise direction. Choose any point as the starting point and call it $0$.
Given $\epsilon>0$, let $k$ be the smallest integer greater than $\frac{1}{\epsilon}$. Cover the circle by $k$ open arcs of length $\epsilon$. Now apply $R_\alpha$ to $x$ again and again just as many times as you want(greater than $k$ of course). You get more than $k$ points inside $k$ arcs. Here you use the pigeon-hole principal to conclude that there are two points within $\epsilon$ arc length to each other. Suppose these two points are $R^n_\alpha x$ and $R^m_\alpha x$ then $R^{n-m}_\alpha x$ lies between $0$ and $\epsilon$. Applying $R^{n-m}_\alpha$ repeatedly, you can obtain a point from the orbit of $x$ in any given open arc of length $\epsilon$. Since $\epsilon$ was arbitrary, it shows that the orbit is dense.
I hope it's clear as to where does the irrationality of $\alpha$ come into play. Essentially, it implies that applying $R_\alpha$ again and again gives a new point each time which makes the pigeon-hole principal applicable. Also welcome corrections in case there are mistakes.
