Given a circle with radius 1. Take an arbitrary starting point. Then go around the circle an infinite number of times, always drawing a point when you are one unit further. So the next point is always an arc length 1 step further. Easy to prove that all points will be different! But my question is: is each point randomly close approached? I believe this is true, but I can’t prove it!

As mentioned in the comments, this has to do with $\mathrm{\pi <!--\; \pi \; -->}$ being irrational. Let's assume that, and go on to prove that you'll end up arbitrarily close to any given point on the circle.

Represent the circle as the interval $[0,2\mathrm{\pi <!--\; \pi \; -->})$ -- basically, a point is represented by how far counterclockwise you have to go from your starting point to get there. In your process, you're starting at 0 and adding 1 to your location every step, and when the location becomes greater than $2\mathrm{\pi <!--\; \pi \; -->}$, you subtract $2\mathrm{\pi <!--\; \pi \; -->}$ so that it lies between 0 and $2\mathrm{\pi <!--\; \pi \; -->}$.

Assume that there's some $0\le <!--\; \le \; -->x<2\mathrm{\pi <!--\; \pi \; -->}$ and some $\mathrm{ϵ<!--\; ϵ\; -->}>0$ for which no point in the interval $(x-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->},x+\mathrm{ϵ<!--\; ϵ\; -->})$ is ever reached. Now, consider the first
$n=\lceil \frac{\mathrm{\pi <!--\; \pi \; -->}}{\mathrm{ϵ<!--\; ϵ\; -->}}\rceil$
steps, and say the locations reached (including the starting point 0) are
$0=x_0<x_1<x_2<\cdots <!--\; \cdots \; --><x_n<2\mathrm{\pi <!--\; \pi \; -->}$
(note that $x_i$ is not the location after i steps, but the ith smallest location in $(0,2\mathrm{\pi <!--\; \pi \; -->})$ that you reach). Since they're all in $[0,2\mathrm{\pi <!--\; \pi \; -->})$, there can't be a distance of greater than $2\mathrm{ϵ<!--\; ϵ\; -->}$ between each pair -- otherwise,
$2\mathrm{\pi <!--\; \pi \; -->}>x_n=(x_n-<!--\; -\; -->x_{n-<!--\; -\; -->1})+(x_{n-<!--\; -\; -->1}-<!--\; -\; -->x_{n-<!--\; -\; -->2})+\cdots <!--\; \cdots \; -->+(x_2-<!--\; -\; -->x_1)+(x_1-<!--\; -\; -->x_0)\ge <!--\; \ge \; -->2n\mathrm{ϵ<!--\; ϵ\; -->}\ge <!--\; \ge \; -->2\mathrm{\pi <!--\; \pi \; -->}.$
So, there must be some integers $a<b<n$ for which the ath step and the bth step are very close to each other -- within $2\mathrm{ϵ<!--\; ϵ\; -->}$. So, the (b−a)th step is within $2\mathrm{ϵ<!--\; ϵ\; -->}$ of 0, since going b−a units forward from the ath location gets you to a point within $2\mathrm{ϵ<!--\; ϵ\; -->}$ of the ath location. Let this location be $\mathrm{\delta <!--\; \delta \; -->}$ units away from 0, so it is either $\mathrm{\delta <!--\; \delta \; -->}$ or $2\mathrm{\pi <!--\; \pi \; -->}-<!--\; -\; -->\mathrm{\delta <!--\; \delta \; -->}$.
Finally, consider the points reached after k(b−a) steps for various integers k. As you add 1 to k, this point moves $\mathrm{\delta <!--\; \delta \; -->}$ units around the circle, either clockwise (if the b−ath location is $\mathrm{\delta <!--\; \delta \; -->}$ units clockwise from 0) or counterclockwise (if it's counterclockwise from 0). Because $\mathrm{\delta <!--\; \delta \; -->}<2\mathrm{ϵ<!--\; ϵ\; -->}$, you can't ever "jump over" the interval $(x-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->},x+\mathrm{ϵ<!--\; ϵ\; -->})$ in this way -- you'll hit some point inside it at step $k(b-<!--\; -\; -->a)$ for some integer $k$. So, there is some step that drops you in $(x-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->},x+\mathrm{ϵ<!--\; ϵ\; -->})$, as desiblack.