Here is an elementary proof for the equidistribution of orbits of irrational rotations. Let $\alpha$ be any irrational number and fix $0\le a<b\le 1$. We need to show that

\begin{equation}\tag{1}\label{eq:equidistribution}
\frac{\# \left\{0\le j < n\,:\,\{j\alpha\} \in[a,b]\right\}}{n}\xrightarrow[n\to\infty]{}b-a.
\end{equation}

The idea is to choose a rational number $p/q$ close enough to $\alpha$ with a large enough denominator, then take $n$ large enough and use the facts that the orbits of a rotation by $p/q$ are evenly spaced within the unit interval, with a distance of $1/q$ between orbit points, and that orbits of the rotation by $\alpha$ stay close to orbits of the same point under rotations by $p/q$ along pieces of orbits of size smaller then $q$. In this respect the current proof is motivated by and similar to the proof by Hardy and Wright mentioned on this page.

We will need the following standard and easy-to-prove lemma:

**Lemma:**
Assume $\alpha\in\mathbb{R}$ and that $p$ and $q$ are coprime integers satisfying
\begin{equation}\tag{2}\label{eq:best}
\left|\alpha-p/q\right| < \frac{1}{q^2}.
\end{equation}
Then for every $0\le i < q$ there exists a unique $0 \le j < q$ such that $\{j\alpha\}\in \left[\frac{i}{q},\frac{i+1}{q}\right)$.

**Proof:**
Without loss of generality assume that $\alpha > \frac{p}{q}$. Then $0 < j\alpha - j\frac{p}{q} < \frac{1}{q}$ for every $0\leq j<q$. Writing $jp = s_j q + r_j$ with $0\leq r_j<q$ we see that the integer part of $j\alpha$ is $s_j$ which implies $\frac{r_j}{q} < \{j\alpha\} < \frac{r_j+1}{q}$. Since $p$ and $q$ are coprime the map $j\mapsto r_j$ is a bijection on $\{0,\ldots,q-1\}$ so we are done.

Getting back to our sheeps, fix any $\varepsilon>0$. By Dirichlet's theorem there are arbitrarily large rational approximations to $\alpha$ that satisfy \eqref{eq:best}. Choose $p$ and $q$ coprime that satisfy \eqref{eq:best} with $q$ large enough ( such that $\frac{2}{q}<\frac{\varepsilon}{3}$ ), and choose $N$ to be large enough ( such that $\frac{q}{N}<\frac{\varepsilon}{3}$ ).

Then, for any $n\ge N$, write $n = sq+r$ with $0\le r < q$. The following estimate holds:
\begin{align*}
\# \left\{0\le j < n\,:\,\{j\alpha\} \in[a,b]\right\} & \ge
\sum_{t=1}^{s} \# \left\{(t-1)q\le j < tq\,:\,\{j\alpha\} \in[a,b]\right\} \\
& \ge s\left(q(b-a) - 2\right) \\
& = n\left(b-a\right) - 2s - r(b-a).
\end{align*}

In the second inequality the number of subintervals from a partition of the unit interval into $q$ subintervals of length $1/q$ that intersect $[a,b]$ bounds the number of orbit points in $[a,b]$. The error in this count is at most two, the reason being that in any partition of the interval into subintervals there are at most two subintervals that intersect $[a,b]$ but are not contained in it. Since $\frac{s}{n} \leq \frac{1}{q} < \frac{\varepsilon}{3}$ and $\frac{r}{n} \leq \frac{q}{n} < \frac{\varepsilon}{3}$, this proves that

$$
\frac{\# \left\{0\le j < n\,:\,\{j\alpha\} \in[a,b]\right\}}{n} \ge b-a - \varepsilon.
$$

The other inequality is proved similarly:
\begin{align*}
\# \left\{0\le j < n\,:\,\{j\alpha\} \in[a,b]\right\} & \le
\sum_{t=1}^{s+1} \# \left\{(t-1)q\le j < tq\,:\,\{j\alpha\} \in[a,b]\right\} \\
& \le (s+1)\left(q(b-a) + 2\right) \\
& = n\left(b-a\right) + 2(s+1) + (q - r)(b-a),
\end{align*}
so,
$$
\frac{\# \{0\le j < n\,:\,\{j\alpha\} \in[a,b]\}}{n} \le b-a + \varepsilon.
$$
Since $\varepsilon$ is arbitrary, this proves \eqref{eq:equidistribution}.