Homeomorphism of the unit disk fixes the unit circle I'm trying to solve this problem

Suppose $f : D \rightarrow D$ is a homeomorphism, then $f(S^1)=S^1$. 

I really dont know how to tackle this problem , i started trying to prove it by contradiction  but i dont  know how to go from there. Any tips are appreciated, just want a few advices to get the exercise started.Thanks in advance.
 A: Assume $f(S^1) \ne S^1$. Then $f(S^1) \not\subset S^1$ or $S^1 \not\subset f(S^1)$. It suffices to consider the case $f(S^1) \not\subset S^1$ because in the other case we have $f^{-1}(S^1) \not\subset S^1$ and we can consider the homeomorphism $f^{-1}$ instead of $f$.
Some $x_0 \in S^1$ is mapped to $y_0 = f(x_0) \in \Delta = D \setminus S^1$. Hence $f$ establishes a homeomorphism $f' : D \setminus \{x_0\} \to D \setminus \{y_0\}$. But $D \setminus \{x_0\}$ is contractible (radial contraction to $0$) and $D \setminus \{y_0\}$ is homotopy equivalent to $S^1$, thus the two spaces cannot be homeomorphic (consider e.g. fundamental groups).
A: A more elementary proof uses homotopies. That is, you  can also get a contradiction by observing that homeomorphisms preserve simple-connectedness: if $f:X\to Y$ is a homeomorphism, and $X$ is simply connected, then it's easy to show that $Y$ is path connected, so fixing a point $y_0\in Y$, and taking any loop $\ell:I\to Y$ at $y_0$, we have that $f^{-1}\circ \ell$ is a loop in $X$, and since $X$ is contractible, there is a homotopy $H$ from $f^{-1}\circ \ell$ to $f^{-1}(y_0)$. Then, $f\circ H$ is a homotopy from $\ell$ to $y_0$, so $Y$ is contractible. 
