an homeomorphism from the plane to the disc

I find the map that just radially shrinks the plane into the disc more natural. Explicitly, this is (after identifying $\mathbb{C}$ with $\mathbb{R}^2$), $$(x,y)\mapsto \frac{1}{\|(x,y)\|+1} (x,y)$$

Let $f\colon\mathbb R^{\ge0}\to[0,1)$ be an order preserving bijection (this means that $f$ is a homeomorphism, since both sets have the order topology).

For $z\in\mathbb C$ we can write $z=\rho e^{i\theta}$. Map $z\mapsto f(\rho) e^{i\theta}$.

For example, $f(x)=\dfrac{x}{1+x}$ is continuous on $\mathbb R^{\ge 0}$, it is strictly increasing (note that we have $f'(x) = \dfrac{1}{(1+x)^2}>0$ everywhere) and $\displaystyle\lim_{x\to\infty} f(x) = 1$, as wanted.

Therefore the map $\rho e^{i\theta}\mapsto \dfrac{\rho}{1+\rho}e^{i\theta}$ should do the trick.