Homeomorphism Compact Subsets Are there compact subsets $A,B \subset \mathbb{R^2}$ with $A$ not homeomorphic to $B$ but $A \times [0,1]$ homeomorphic to $B \times [0,1]$?

Yes. Consider the two sets below.

Edit: Jon, let me explain in more detail why the two sets (call them $A$ and $B$, respectively) are not homeomorphic. I find in practice that showing two sets to not be homeomorphic is a bit cumbersome. I didn't fill out all of the details for that reason, leaving them to you; let me know if you would like more explanation for anything.
Suppose, seeking a contradiction, that $h:A\to B$ is a homeomorphism.
Let $\alpha:[0,1]\to B$ be a path between
the two points where the crosses intersect the edge of the square
and let $\beta:[0,1]\to B$ be a path between the two points where the sticks intersect the edge. Choose $\alpha$ and $\beta$ so that their images do not intersect, that is, so that $\alpha([0,1])\cap\beta([0,1])=\varnothing$. Now consider the paths $\alpha'=h^{-1}\circ\alpha:[0,1]\to A$ and $\beta'=h^{-1}\circ\beta:[0,1]\to A$. It is still true that $\alpha'$ is a path between the intersections of the crosses and the squares, and that $\beta$ is a path between the intersections of the sticks and the squares (why?). But the images of $\alpha'$ and $\beta'$ must intersect (why?), and that is a contradiction.
As for your second question -- why $A'\times[0,1]$ and $B'\times[0,1]$ are not homeomorphic -- let me give you some intuition. The sets $A\times[0,1]$ and $B\times[0,1]$ are homeomorphic because you can move the "branches" across the top of the cube contained in $A\times[0,1]$. But the space $A'\times[0,1]$ is $A\times[0,1]$ with the interior, top, and bottom of the cube removed -- there is no more "top" to move the branches across, and a homeomorphism won't work.
