These two spaces are not homeomorphic...right? why is $\Bbb R\times[0,1]\not \cong \Bbb R^2$? we can't use the popular argument of deleting a point and finding that one has more path components than the other here.
So my idea is to delete a strip $\{0\}\times[0,1]$ from $\Bbb R\times[0,1]$.
But is $\Bbb R^2-f(\{0\}\times[0,1])$ always path-conneted when $f:\Bbb R\times[0,1]\to \Bbb R^2$ is a homeo?

There are already more elementary answers given, but the Invariance of Domain theorem immediately shows they are not homeomorphic, as your strip is not open in $\mathbb{R}^2$.
