How to prove that $\arg: S^1\setminus\{-1\}\to (-\pi,\pi)$ is continuous? (With restrictions.) 
How to prove that $\arg: S^1\setminus\{-1\}\to (-\pi,\pi)$ is continuous?

Restrictions: for this proof we cant use logarithms, derivatives or integrals.
I know that $\arg^{-1}$ is bijective and continuous, hence $\arg$ is bijective, but I dont have a clue about how to prove it continuity.
The argument function is defined as the map $e^{i\alpha}\mapsto \alpha$. I thought about define the map $e^{i\alpha}\mapsto e^\alpha$, but it continuity is not more clear than the previous map (the idea behind is that for $e^\alpha$ I can use it strict monotonicity to prove that $e^\alpha\mapsto\alpha$ is continuous).
Other idea is try to show that $\arg^{-1}(\Bbb B(x,\delta))=\Bbb B(e^{ix},\epsilon)$, i.e. $\arg^{-1}$ is open, but I dont have a clue about how to prove this formally. Some help will be appreciated, thank you.

UPDATE:
Attempted proof. We choose some closed interval of the kind $C_n:=[-\pi+\frac1n,\pi-\frac1n]$. Then the restriction of $\arg^{-1}$ to $C_n$ and it image is homeomorphic i.e. $\arg$ is continuous in $\arg^{-1}(C_n)$, because every closed set in $C_n$ is compact and hence it image is compact, what is a definition of continuity for $\arg$.
Because this is true for all $n\in\Bbb N$ then it is true for the entire domain $(-\pi,\pi)$.$\Box$
 A: Suggestion: The inverse map $f:(-\pi, \pi) \to S^{1} \setminus\{-1\}$ defined by $f(t) = \exp(it)$ is a continuous bijection, so is a homeomorphism when restricted to an arbitrary closed interval $[a, b] \subset (-\pi, \pi)$, and every point of $S^{1} \setminus\{-1\}$ lies in the interior of the image of some compact set.
A: Using an $\epsilon - \delta$ argument:
Let $\alpha_0\in (-\pi,\pi)$ and $0 < \epsilon < 1$. Choose $\delta = 2 \sin(\frac{\epsilon}{2})$. Then $0 < \delta < 1$. For all $\alpha \in (-\pi,\pi)$, $\lvert e^{i\alpha} - e^{i\alpha_0}\rvert < \delta$ implies $2\sin(\frac{\lvert \alpha - \alpha_0\rvert}{2}) < \delta$. Indeed, $$\lvert e^{i\alpha} - e^{i\alpha_0}\rvert = \lvert e^{i(\alpha+\alpha_0)/2}[e^{i(\alpha - \alpha_0)/2} - e^{-i(\alpha-\alpha_0)/2}]\rvert = \lvert e^{i(\alpha-\alpha_0)/2} - e^{-i(\alpha-\alpha_0)/2}\rvert = 2\left\lvert \sin\frac{\lvert \alpha-\alpha_0\rvert}{2}\right\rvert$$ and since $\lvert \alpha - \alpha_0\rvert < 2\pi$, then $$2\left\lvert \sin \frac{\lvert \alpha - \alpha_0\rvert}{2}\right\rvert = 2\sin \frac{\lvert \alpha - \alpha_0\rvert}{2}$$ Since $0 \le \sin(\frac{\lvert \alpha - \alpha_0\rvert}{2}) < \frac{\delta}{2} < 1$, then $\frac{\lvert \alpha - \alpha_0\rvert}{2} < \sin^{-1}(\frac{\delta}{2})$. Hence, $\lvert \alpha - \alpha_0\rvert < 2\sin^{-1}\left(\frac{\delta}{2}\right) = \epsilon$. Since $\alpha_0$ was arbitrary, the $\arg$ is continuous.
