This question was posted on twitter here as a quiz but the author never gave an answer, so I thought I'd try here. I don't have much experience with topology so I'm stumped. From searching online it seems that it's not the case in general (e.g. $X \times \mathbb{R} \cong \mathbb{R}^n$ does NOT imply $X \cong \mathbb{R}^{n-1}$) so I figure that the answer will have to use the fact that the product space is two dimensional, or that the quotient (either the subspace of $\mathbb{R}^2$ homeomoprhic to $X$ or the fibers of that map) are one dimensional. Some people on twitter mentioned trying to prove that $X$ (or the image of $X$ embedded in $\mathbb{R}^2$ has all the properties that distinguish the real numbers line topologically (i.e. connected, locally connected separable metric space, such that every point is a strong cut point). Perhaps once can also impose an ordering on $X$ that matches the ordering of $\mathbb{R}$?
I'm pretty sure that the map $ \mathbb{R}^2 \rightarrow X $ induced by the product is a quotient map, perhaps this preserves useful properties such as being connected, locally connected, etc.? Also $X$ seems to be a sub-metric space of $\mathbb{R}^2$ by embedding every point $x \in X$ into $\phi(x, 0) \in \mathbb{R}^2$ where $\phi$ is the isomorphism from $X \times \mathbb{R}$ into $\mathbb{R}^2$.
Any help would be appreciated, thanks!
Yes, $X$ will be homeomorphic to ${\mathbb R}$. Indeed, $X\times \{0\}\subset {\mathbb R}^2$, hence, $X$ is metrizable. Since $X$ is a quotient of ${\mathbb R}^2$, it follows that $X$ is separable. For the same reason, it is connected. Furthermore, for every $x\in X$, $$ L_x=\{x\}\times {\mathbb R}\subset {\mathbb R}^2 $$ is properly embedded, hence, by Jordan Curve Theorem, it separates ${\mathbb R}$ in exactly two path-components. Hence, $\{x\}$ separates $X$ in exactly two path-components.
It, therefore, follows from the 2nd answer here that $X$ is homeomorphic to the real line.