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?
 A: That spaces have different one-point (Alexandroff) compactifications, hence they cannot be homeomorphic. 
The one-point compactification of $\mathbb{R}^2$ is the sphere $S^2$, while the one-point compactification of $\mathbb{R}\times[0,1]$ is a closed disc in $\mathbb{R}^2$ with a pair of its boundary points identified. In $S^2$, two open neighbourhoods of two different points are always isomorphic, and by removing any point in a neighbourhood it stays connected. In the last space, there is a point $u$ with an unusual behaviour: by taking an open neighbourhood $U$ of $u$, $U\setminus\{u\}$ is disconnected.
A: The property of simple connectivity will distinguish between $\Bbb{R} \times[0,1]$ and $\Bbb{R}^2$. When we remove one point from $\Bbb{R} \times[0,1]$ then it is simply connected but removing one point from $\Bbb{R}^2$ then it is not simply connected.
A: 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.
A: 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 $\mathbb{R}^2$.
A: Here’s another variation on the idea of using compactifications to avoid algebraic topology: $\Bbb R\times[0,1]$ has a two-point compactification, but $\Bbb R^2$ does not. 
The two-point compactification of $\Bbb R\times[0,1]$ is pretty evident. Suppose that that $\Bbb R^2$ has a compactification $X=\{p,q\}\cup\Bbb R^2$, where $p\ne q$. Let $U$ and $V$ be disjoint open nbhds of $p$ and $q$; $K=\Bbb R^2\setminus(U\cup V)$ is a compact subset of the plane. Let $G=U\cap\Bbb R^2$ and $H=V\cap\Bbb R^2$; $G$ and $H$ are open in $\Bbb R^2$ and partition $\Bbb R^2\setminus K$.
$K$ is compact, so let $r_0=\max\{\|x\|:x\in K\}$. For each $r>r_0$ let $C_r$ be the circle of radius $r$ centred at the origin. $C_r$ is connected, so for each $r\ge r_0$ we must have $C_r\subseteq G$ or $C_r\subseteq H$. Let $I_G=\{r>r_0:C_r\subseteq G\}$ and $I_H=\{r>r_0:C_r\subseteq H\}$. Each $C_r$ is compact, so $I_G$ and $I_H$ are open subsets of the connected set $(r_0,\to)$; without loss of generality $I_G=(r_0,\to)$ (and $I_H=\varnothing$). It follows that $H$ is a bounded open set in $\Bbb R^2$ and hence that $W=X\setminus\operatorname{cl}_{\Bbb R^2}H$ is an open nbhd of $q$ in $X$. But then $V\cap W=\{q\}$ is open in $X$, and $q$ is isolated, which is impossible.
