Homeomorphism from $\bar{D}(0,1)$ to itself takes $\mathbb{S^1}$ to $\mathbb{S^1}$

Question

I have been trying to proof the following result, but haven't found any progress. Any hint? I'm studying topology from Munkres and Dugundji

Any homeomorphism from the closed disk $\bar{D}(0,1) = \{(x,y)\in \mathbb{R^2}: x^2+y^2\leq 1\}$ to itself sends the unit circle $\mathbb{S^1}$ to $\mathbb{S^1} $ and the open disk ${D}(0,1)$ to $D(0,1)$.

Answer 1

A point $p\in\Bbb S^1$ is characterised by the fact that $\overline{D}(0,1)\setminus\{p\}$ is contractible, whereas $\overline{D}(0,1)\setminus\{p\}$ deformation retracts on $\Bbb S^1$ when $p$ lies in the interior ${D}(0,1)$.