I trying to show that any homeomorphism $h:D^2\to D^2$, where $D^2 =\left\{x\in \mathbb{R}^2: |x|\leq 1\right\}$ takes $S^1$ to $S^1$ and ${D^2}^\circ$ to ${D^2}^\circ$(interior of a disk).
Supposed that $a\in {D^2}^\circ$ and $h(a) \in S^1$. Then, let $A = {D^2}-\left\{ a\right\}$ and $B = D^2-\left\{ h(a)\right\}$. Hence $h:A\to B$ is again a homeomorphism. We know that $h$ induce a isomorphism on the fundamental group of $\pi_1(A)$ onto $\pi_1(B)$. But $\pi_1(A)$ is isomorphic to $\mathbb{Z}$. My intuition says that $\pi_1(B)$ is trivial. If we define a deformation retract $F(s,t) = x_0 (1-t) + x t$ where $x_0$ is a point of $S^1$ and $x\in B$ then I believe that this is a deformation retract of $B$ to $x_0$. But the question is, what's happens if $x_0$ is very close to $h(a)$? $F$ is well defined if $B $ is convex, but this set if really a convex set?
It is convex. Assume that $v,w\in D^2$, $v\neq w$. Then $(1-t)v+tw\in S^1$ if and only if $t=0$ and $v\in S^1$ or $t=1$ and $w\in S^1$. And therefore removing points from boundary does not affect convexity. Because boundary points are never interior points of an interval.
Note however that you do not need full power of convexity. It is enough to note that it is a star set. Even more, just pick $x_0=0$ and consider renormalization $F(x,t)=tx$. It is a lot simpler to show that $a$ cannot be in the image of $F$, and therefore $F$ is well defined.