Prove that the orbit of an iterated rotation of 0 (by (A)(Pi), A irrational) around a circle centered at the origin is dense in the circle.

I think the title of the question says it all. I unfortunately did not seem to conclude anything. Some ideas I had:

It is easy to show that (given $T$ is the rotation) $\{T^n(\theta)\}$ is a set of distinct points. Furthermore, given that the circle is a compact metric, it must have a limit point $x$. By continuity of the rotation function, $T^n(x)$ is a limit as well since taking $T$ of every term yields the same sequence (with only the first term removed). By induction, we have infinitely many distinct limit points $\{T^n(x)\}$.

That's all I could come up with! It was also easy to show that the orbit is infinite. I still don't seem to be able to get close to the required result however.

First, let's get rid of the $\pi$ and talk about fractions of a circle and regular irrational numbers instead.

Let $a<1$ be an irrational number, and assume that the orbit that you get by going $a$ of the circle (say, clockwise) $n$ times around a point is not dense in the circle, that is, there exists an open interval $I$ on the circle of size $\varepsilon$ which it does not cross.

Assume there were $m>n$, such that the points corresponding to rotations by $na$ and $ma$ have a distance $d<\varepsilon$. This would mean that rotation by $m-n$ gives a new point, which is of distance $d$ from the previous point, and because $d<\varepsilon$, this means that if this rotation is done repeatedly, eventually a point will be in $I$.

So there can be no such $m,n$, meaning that points in the orbit are at least $\varepsilon$ apart, which means that there are at most $\frac{1}{\varepsilon}<\infty$ of them, contradicting the fact that $a$ is irrational.