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?

To riff off Jack D'Aurizio's idea to use compactifications (and trying to avoid Algebraic Topology notions...):
The Cech-Stone compactification of $\mathbb{R}^2$ has a connected remainder, while that of $\mathbb{R} \times [0,1]$ has two components in its remainder. The proof is similar to that of the Cech-Stone compactification of $\mathbb{R}$.
The Freudenthal compactification should have the same properties for its remainder as well.
