There is a really easy prof that just uses the Pigeonhole principle. Let $\alpha$ be irrational.

**Lemma:** For any $\delta >0$, there is an $n>0$ such that $(n \alpha) \in (- \delta, \delta) \setminus \{ 0 \}$.

**Proof:** Choose $N$ large enough that $1/N < \epsilon$. Divide $[0,1]$ up into $N$ segments of length $1/N$. By pigeonhole, there are $j$ and $k$ such that $(j \alpha)$ and $(k \alpha)$ land in the same segment. So $((j-k) \alpha) \in (-\delta, \delta)$. Since $\alpha$ is irrational, $((j-k) \alpha) \neq 0$. $\square$

Now, fix $0 \leq p < q \leq 1$ and $\epsilon>0$. Our goal is to show that
$$q-p-\epsilon \leq \lim_{N \to \infty} inf \frac{\#\{ k \leq N : (k \alpha) \in (p,q) \}}{N} \leq \lim_{N \to \infty} sup \frac{\#\{ k \leq N : (k \alpha) \in (p,q) \}}{N} \leq q-p+\epsilon.$$

Choose $\delta >0$ such that
$$\frac{(q-p)/\rho -1}{1/\rho+1} \geq q-p-\epsilon \quad \mbox{and} \quad \frac{(q-p)/\rho +1}{1/\rho-1} \leq q-p+\epsilon$$
for all $\rho$ with $|\rho| < \delta$.

Choose $n$ such that $n \alpha \in (-\delta, \delta) \setminus \{ 0 \}$. Write $\rho = (n \alpha)$.
Break up the set of values $(k \alpha)$ up into arithmetic progressions based on $k$ modulo $n$. So each segment is of the form $(\beta + j \rho)$. It is enough to prove that the lim inf and lim sup contributed by each progression lie between $q-p-\epsilon $ and $q-p+\epsilon$, as the total contribution is a weighted average from the contributions from the progressions.

Break each progression up into segments according to $\lfloor \beta + j \rho \rfloor$. The initial segment and final segment each contain at most $1/ |\rho|$ terms. The segments in the middle contain between $1/\rho - 1$ and $1/\rho+1$ terms of which between $(q-p)/\rho -1$ and $(q-p)/\rho+1$ are between $p$ and $q$. Since $((q-p)/\rho-1)/(1/\rho+1)$ and $((q-p)/\rho+1)/(1/\rho-1)$ were chosen to lie in $(q-p-\epsilon, q-p+\epsilon)$, the average of all of these terms lies in the required interval. And the terms from the initial segments can only drag the average off by at most $(2/|\rho|)/(N/n)$, which goes to $0$. QED.