First, it's very easy to show this "the discrete way": Let $r$ be real and assume $r \ne 0.$ If $r > 0$, then $\displaystyle{\frac{r}{1 + r} \in (0, 1) \text{ and } f\left(\frac{r}{1 + r}\right) = r}$. The other case is just as easy.

On the other hand, the proof in quotes below is a bit confusing as I don't know calculus yet. I do know how to work with the rigorous definition of limit, but that's no help here. Does the red part in the proof below imply the blue part or do the colored parts together imply the conclusion? Can somebody, please, refer me to (or elaborate) theorems and definitions used in the proof in quotes? Thanks.

> 
> The function $f$ is defined on $(-1, 1)$. Also, $\color{red}{f(0) = 0, f(x) > 0 \text{ when } 0 < x < 1, \ f(x) < 0 \text{ when } -1 < x < 0}$. This function also has the property that $\color{red}{\displaystyle{\lim\_{x \to 1^-}f(x) = +\infty, \ \lim\_{x \to -1^+} = -\infty}}$. If you recall enough information about continuous functions from calculus, you might see that $\color{blue}{\text{ this function is continuous on the interval $(−1,1)$}}$. From this information, it
>  follows that $f((−1,1)) = \mathbb R$ and that $f$ is onto.
> 
> 
>