Homeomorphism between $X$ and $X\times \mathbb R$ This question comes as a result of me answering The graph of a real valued continuous function on a metric space is homeomorphic to what set? which got me thinking.

Is it possible to have two topological spaces $X$ and $Y$ such that $X\times Y$ is homeomorphic to $X$?

The answer to this question is clearly (correct me if I'm wrong) yes, since any set $X$, equipped with either the discrete or the trivial topology, is homeomorphic to $X\times X$.

So, the next question would be

Is it possible to have a topological space $X$ such that $X\times \mathbb R$ is homeomorphic to $X$?

And here, taking $X=\emptyset$ gets us through. However, the question I don't know how to answer is

Is it possible to have a nonempty topological space $X$ such that $X\times \mathbb R$ is homeomorphic to $X$?

I don't see an easy answer to this. The "simple" answers don't seem to work here, and I'm out of ideas.
 A: For a non-trivial example for the first Q (with $Y=X$): If $S$ is a non-empty space and $X$ is the Tychonoff product  of infinitely many copies of $S$ then $X$ is homeomorphic to $X\times X.$
A: 
The answer to this question is clearly (correct me if I'm wrong) yes, since any set $X$, equipped with either the discrete or the trivial topology, is homeomorphic to $X\times X$.

Let me answer that: if $X$ is finite and has at least two elements, then no matter what topology you choose, $X$ will never be homeomorphic to $X \times X$, simply because there isn't even a bijection between them (much less a homeomorphism). If $X$ is infinite, however, then there is a bijection $X \cong X \times X$, and this does induce a homeomorphism when $X$ is equipped with either the discrete or the trivial topology. Same if $X$ is either empty or a singleton.
A: If $X = Y^\kappa$ is any infinite product of copies of $Y$, then $X \times Y \cong X$, and $X \times X \cong X$.
The Cantor space $\{0,1\}^\omega$ is such an example.
A: Yes. For instance $\Bbb R^\infty$.
