$\mathbb{R}\cong X\times Y$ (homeomorphic) implies $X$ or $Y$ is a point. The following is Problem 2.1.1 from Tammo tom Dieck's Algebraic Topology:

Suppose $\mathbb{R}\cong X\times Y$ (homeomorphic). Then $X$ or $Y$ is a point.

FWIW, the section 2.1 is about (path) connected components (i.e., $\pi_0$) and the notion of homotopy.
I tried to use the usual trick by removing one point and consider the number of components, but failed. The spaces $X,Y$ are arbitrary, and so I don't know how to handle this.
Any hints will be appreciated!
 A: This comes from the following theorem:
Theorem: Suppose $X$ and $Y$ are path connected and each have at least two points.  Given any point $(x_0, y_0) \in X \times Y$, the space $X \times Y -\{(x_0,y_0)\}$ is still path connected. 
In your case, $X \times Y$ must be path connected since $\mathbb{R}$ is.  Hence, each of $X$ and $Y$ are path connected, so the above result is applicable.  But upon removing a point we get into trouble, since $\mathbb{R}-pt$ is no longer path connected, but $X \times Y -\{(x_0,y_0)\}$ is.  So all you have to do is believe in the theorem.
Sketch Proof.  Let $(a,b)$ and $(c,d)$ be any two points in $Z=X \times Y -\{(x_0,y_0)\}$.  We must show there exists a path between these two points, the path living in $X \times Y -\{(x_0,y_0)\}$ (so, avoiding $(x_0,y_0)$).  The argument comes in several cases.
First, we could have $a=c\neq x_0$.  Use the fact that $\{a\} \times Y\cong Y$ is path connected and build the path in that slice.  That clearly gives a path in $Z$.  The same argument works if $b=d\neq y_0$.
Next, the case $a=c=x_0$.  It follows that $b \neq y_0$ and $d \neq y_0$. Here's where you need the assumptions that $|X|, |Y| \geq 2$.   We can choose a point $x'\in X$ which is different from $x_0$.  Now create the needed path in three segments:
$$
(a,b) \to (x',b) \to (x',d) \to (c,d)
$$
working in the obvious coordinate slices in each leg (see the argument for the first case).  This builds a path $(a,b) \to (c,d)$ in $Z$. The same thing works if $b=d=y_0$.
Finally, there is the case $a \neq c$ and $b \neq d$.  Construct two different three-legged paths
$$
(a,b) \to (c,b) \to (c,d)
$$
and
$$
(a,b) \to (a,d) \to (c,d).
$$
It is not hard to check that $(x_0,y_0)$ cannot live on both of these, so one of these gives the escape. Hence we're done. 
A: Hint: By assumption, $X\times Y$ is path-connected, hence so are $X$ and $Y$. Now you should be able to proceed with your usual trick, e.g. by assuming that both $X$ and $Y$ have at least two distinct points and then proving that $X\times Y$ minus a point must still be path-connected.
