$H: \mathbb{R^2 } \rightarrow \mathbb{R^2}$ be a homeomorphism such that $H(\partial B_1)= \partial B_2$. Show that $H(B_2)=B_1$. Let $B_1,\ B_2 \subset  \mathbb{R^2}$ be disks, and let $H:  \mathbb{R^2 } \rightarrow  \mathbb{R^2}$ be a homeomorhism such that $H(\partial B_1)= \partial B_2$. Show that $H(B_2)=B_1$.
My attempt:
Since $\partial B_1$, and $\partial B_2$ are 1-spheres, by Schönflies Theorem there exist homeomorphisms $H_1,H_2: \mathbb{R^2} \rightarrow \mathbb{R^2}$ such that $H_1(S^1)= \partial B_1$, and $H_2(S^1)= \partial B_2$.
So, $S^1 = H_i^{-1}(\partial B_i)$ for $i=1,2$.
$$H_1^{-1}(\partial B_1)=H_2^{-1}(\partial B_2)$$
$$H_2 \circ H_1^{-1}(\partial B_1) = \partial B_2$$
It is clear that $H_2 \circ H_1^{-1}$ is a homemorphism. Now, I wish to prove the following claim.
Claim: $H_2 \circ H_1^{-1}(B_1)=B_2$.
This is all I have done so far. Am I on the right track? I am open to any help.
Additionally, even though it seems a bit irrelevant I have another question. If we have a homeomorphism $H: \mathbb{R^n} \rightarrow \mathbb{R^n}$, and we have two homeomorphic subsets of $\mathbb{R^n}$, say $B$ and $D^2$,(unit disk), can we say directly that $H: B \rightarrow D^2$ is also a homeomorphism?

Hint: (a very simple special case of) the Jordan curve theorem tells you that both $\partial B_1$ and $\partial B_2$ separate the plane into an "inside" (bounded and connected) region (say $I_1$ for $\partial B_1$ and $I_2$ for $\partial B_2$) and an "outside" (unbounded and connected) region (say $O_1$ for $\partial B_1$ and $O_2$ for $\partial B_2$). Under your homeomorphism, a simple argument using connectivity says that you must have $H(I_1) = I_2$ or $H(I_1) = O_2$. But the latter case is impossible, because it implies that the compact set $H(I_1 \cup \partial B_1)$ is homeomorphic to the non-compact set $O_2 \cup \partial B_2$. So $H(B_1) = H(I_1 \cup \partial B_1) = I_2 \cup \partial B_2 = B_2$.
