Winding semi-circle? Let us say that $F : \Bbb R → S^1$ is defined by, 
$F(t) = exp(2πti) = cos(2πt) + isin(2πt)$
Say that I am given some open arc in $S^1$, defined by the set:
$C(x) = \{y ∈ S^1 | d(x, y) < 0.01\}$ , where $d(x, y) = |x-y|$ is your typical Eudclidean distance. 
This seems to be smaller than a semi-circle, right? 
What I do not get is how $F^{-1}(C)$ has a countably infinite number of open and disjoint intervals and that the restriction of $F$ to any of those open and disjoint intervals is a homeomorphism from the interval onto $C$. Is there some proof of this? Thanks!
 A: First of all note that $F$ restricted to $[r, r+1)$ is a continuous bijection for any $r\in\mathbb{R}$. Which I leave as an exercise. Moreover $F$ restricted to $(r,r+1)$ is a homeomorphism onto image (being $S^1$ without a point). The inverse can be reconstructed from the complex argument.
Now fix a point $q\in S^1$. Assume that $C\subseteq S^1\backslash\{q\}$. This is your case due to "$<0.01$" condition. Also fix $p\in F^{-1}(q)$.
For any $k\in\mathbb{Z}$ let $G_k:(p+k,p+k+1)\to S^1\backslash\{q\}$ be given by $G_k(t)=F(t)$. So again it's a restriction. As I've mentioned earlier $G_k$ is a homeomorphism. Furthermore
$$F^{-1}(C)=\bigcup_{k\in\mathbb{Z}} G_k^{-1}(C)$$
Finally each $G_k^{-1}(C)$ has to be an interval when $C$ is connected. It has to be open when $C$ is. Both conditions are satisfied in your case. And $\{G_k^{-1}(C)\}_{k\in\mathbb{Z}}$ are pairwise disjoint by definition regardless of what $C$ is.
Can you fill all the missing details?
A: Not an answer, but I hope this picture can help you develop some intuition. It is directed to you "getting how $F^{-1}(C)$ is a disjoint union of countably many intervals...". 

A: Let me answer your question indirectly by starting with some things that may be obvious to you from your understanding of trigonometry.
Suppose that I fix a nonempty open subinterval $(r,s) \subset \mathbb R$ such that $0 < s-r<1$. Taking the image under $F$ I get a circular arc
$$C = F(r,s) = \{exp(2 \pi t \, i) \mid r < t < s\}
$$
Furthermore, I can see that the restriction $F \mid (r,s)$ is a homeomorphism from $(r,s)$ to $C$: the restriction is injective because of the requirement that $0 < s-r<1$; and the inverse homeomorphism $C \mapsto (r,s)$ can easily be written down with certain inverse trig functions of the $x,y$ coordinates of each point in $C$, as one learns in precalculus. So far so good.
Next, knowing the periodicity of cosine and sine, I deduce that the set $F^{-1}(C)$ is the union of a countably infinite number of open and disjoint intervals of the form $(r + 2 \pi k, s + 2 \pi k)$, $k \in \mathbb Z$. 
Furthermore, each of the restrictions $F \mid (r + 2 \pi k, s + 2 \pi k)$ is a homeomorphism to $C$, because I can again use periodicity of cosine and sign to deduce that this map is the composition of two homeomorphisms: the first homeomorphism is the map from the interval $(r + 2 \pi k, s + 2 \pi k)$ to the interval $(r,s)$ given by subtraction of $2 \pi k$; and the second homeomorphism is $F \mid (r,s)$.
Okay, now back to your question. 
Choose $u \in \mathbb R$ such that $F(u) = x$. 
Next, let $\theta$ be the smallest positive number such that $d(x,  F(u+\theta)) = .01$. 
Let $s = u+\theta$ and let $r = u - \theta$, and observe that $d(x,F(r))=d(x,F(s))=.01$. 
Also observe that the set $F(r,s)$ is exactly your circular arc $C(x)$. 
All of the conclusions that you were asking for have been proved!
