Homeomorphism of the Disk I'm working through Massey's "Basic Course in AT."  One of the problems is prove that a homeomorphism of the closed disk maps the boundary to the boundary and the interior to the interior.  How would one prove this?  I can't seem to get this one problem.

A homeomorphism $\phi:D \to D$ maps open sets to open sets. Pick a point $P$ in the interior of the disk and a disk $D(P,\delta)$ which lies inside the given disk $D$. If $P$ would be mapped on the boundary of $D$ then $\phi(P)$ would not be in the interior of $\phi(D(P,\delta))$; contradiction with the fact that $\phi(P) \in \phi(D(P,\delta))$ and $\phi(D(P,\delta))$ is open. 
This means that interior points are mapped to the interior of the disk. 
Pick now $Q$ on the boundary of $D$. If $\phi(Q)$ is on the interior of $D$, then by a similar argument with the one above, since $\phi^{-1}$ is also a homeomorphism it follows that $\phi^{-1}(\phi(Q))=Q$ is in the interior of $D$. Contradiction. 
Therefore $\phi$ maps the boundary of the disk onto the boundary of the disk.
[edit] As mentioned in the comments, I used the standard euclidean topology, not the induced one, and that might be a problem. I think that the best solution still remains the one mentioned in the first comment: 
That if the homeomorphism $\phi : D\to D$ maps $P$ to $Q$ then $D\setminus\{P\}$ and $D\setminus \{Q\}$ are homeomorphic. If $P$ is on the boundary and $Q$ is in the interior (or the other way around) we get two domains which are not homeomorphic: one is simply connected, one isn't (it has a "hole")
