I want to prove that $F\colon \mathbb{R}\to \mathbb{S}^1$, $\quad\theta\mapsto (\cos\theta,\sin\theta)$ is a local homemorphism.
I reasoned as follows:
1) $F$ is clearly continuous.
2) For every $\theta\in \mathbb{R}$ there is $k\in\mathbb{Z}$ such that $\theta\in (-\pi/2+k\pi,\pi/2+k\pi)$ or $\theta\in (k\pi,\pi+k\pi)$.
2.1) If $\theta\in (-\pi/2+k\pi,\pi/2+k\pi)$, than
$$F\colon (-\pi/2+k\pi,\pi/2+k\pi)\to \mathbb{S}^1\cap \{(x,y)\in \mathbb{R}^2\mid (-1)^kx>0\}$$
is an homeomorphism, with inverse
$$G\colon \mathbb{S}^1\cap \{(x,y)\in \mathbb{R}^2\mid (-1)^kx>0\}\to (-\pi/2+k\pi,\pi/2+k\pi), \quad(x,y)\to \arctan(y/x)+k\pi$$
2.2) If $\theta \in (k\pi,\pi+k\pi)$, then
$$F\colon (k\pi,\pi+k\pi)\to \mathbb{S}^1\cap \{(x,y)\in \mathbb{R}^2\mid (-1)^ky>0\}$$ is an homeomorphism, with inverse
$$G\colon \mathbb{S}^1\cap \{(x,y)\in \mathbb{R}^2\mid (-1)^ky>0\}\to (k\pi,\pi+k\pi), \quad(x,y)\to \operatorname{arccotan}(x/y)+k\pi$$
1) Am I right?
2) Is there a shorter and smarter proof? (But still elementary proof)
3) Is also true that all the above homeomorphisms are diffeomorphisms?
The answer for the three questions is yes. Let me show a alternative prove. Consider the map $f:\mathbb R \times (0,\infty) \to \mathbb R^2 \setminus \{0\}$ given by $f(t,r) = r(\cos(t),\sin(t))$.
Notice this map is smooth and surjective (because the map gives you polar coordinates). The jacobian of this map is the matrix \begin{bmatrix} \cos(t) & -r\sin(t) \\ \sin(t) & r\cos(t) \\ \end{bmatrix} whose determinant is $r>0$. Therefore, by the inverse function theorem this map is local diffeomorphism. The map you want is the restriction $F(t) = f(t,1)$, that will be local diffeomorphism too.