If $\alpha$ is irrational, $\lim_{n \to \infty} \sin(n \alpha \pi)$ DNE

I've read an answer to this on another post here.

According to the green-checked answer there, let $y=x/2$, and we first know that $a_n = \text{ny} \mod{1}$ (i.e., the fractional part of $ny$) is dense on $[0,1)$, but I didn't really get the rest of the proof.

My questions are:

1. Why do we need to define $y = x/2$?
2. How is the density of $a_n$ on interval $[0,1)$ useful here?
3. How is $sin(n\pi x)=sin(2 \pi a_n)$ ???
4. I really didn't get the last part of the proof. We showed that there for all $n_0$, there always exist some $n_1 > n_0$ and $n_2 > n_0$ such that $sin(2 \pi a_{n_1}) \not = sin(2 \pi a_{n_2})$. How does this prove the limit DNE?

Here's the full proof from the post:

For convenience we define the irrational number $y = x/2$.

The sequence $a_n = ny \operatorname{mod} 1$ [is dense][1] on $[0, 1)$.(The [equidistribution theorem][2] gives the stronger result that it's uniformly distributed on that interval, but we don't need that for our proof)

This implies that for every $n_0 \in \mathbb{N}$ and $0 \le a < b \le 1$, you can find an $n>n_0$ such that $a_n \in (a, b)$. (because otherwise, there would only be a finite number of elements in the interval $(a, b)$, and then it wouldn't be dense)

The idea of this proof is that for every $n_0 \in \mathbb{N}$, we can always find an $n > n_0$ such that $\sin(n \pi x) > c_1$, and we can also find an $n > n_0$ such that $\sin(n \pi x) < c_2$, where $c_1 > c_2$.

We take $c_1 = \sin(2 \pi \cdot 0.1)$ and $c_2 = \sin(2 \pi \cdot 0.6)$.

Given an $n_0$, we can find an $n > n_0$ such that $a_n \in (0.1, 0.2)$. We have:

$$n \pi x \operatorname{mod} 2\pi = 2\pi n y \operatorname{mod} 2\pi =2\pi(n y \operatorname{mod} 1) =2\pi a_n \\\sin(n \pi x) = \sin(2\pi a_n) > c_1$$Similarly, for every $n_0$ we can find an $n > n_0$ such that $a_n \in (0.6, 0.7)$ and$$\sin(n \pi x) < c_2$$

So, the sequence $n \to \sin(n \pi x)$ has no limit.