The real line is not homeomorphic to any non-trivial product space I came across a question that interested me recently. It asked the following:

Prove that if $\mathbb R$ is homeomorphic to $X \times Y$, then $X$ or $Y$ is a singleton set.

I have an easy proof using path-connectedness. I was interested if there is an even more elementary argument. The notion of connectedness had not yet been introduced in the text.
Let $X$, $Y$ be topological spaces so that $\mathbb{R}$ is homeomorphic to $X \times Y$. Since $\mathbb{R}$ is connected, clearly so is $X \times Y$. Now $X$ is a surjetive image of $X\times Y$, and so is $Y$. Hence both $X$ and $Y$ are also connected. 

Note now that $\mathbb{R}$- and so $X\times Y$- has the property that complement of any $1$ point subset is not connected. Let us show then that either $X$ and $Y$ contains only one element. Assume the contrary, that $X$ contains at least two elements $x_1$, $x_2$, and $Y$ contains $y_1$, $y_2$. Let us show that in fact the space $Z\colon =X \times Y \backslash\{ ( x_1, y_1)\}$ is connected. For this it enough to notice that any point $(x,y)$ of $Z$ is in the same connected component of $Z$ with $(x_2, y_2)$. Indeed, any $(x,y)$ with $x\ne x_1$ is contained together with $(x, y_2)$ in the connected subset $\{x\} \times Y$ of $Z$. 

Note: What we showed in fact was that any connected space such the complement of some ( thanks @bof) point is not connected, is not a product space in a non-trivial way.
UPDATE 2: Ok, I realize critical assertion that $\pi_X \circ h \colon \mathbb{R} \to X$ is a homeomorphism was wrong. Going to see if I can stab at it without connectedness when I wake up.

UPDATE 1: Without loss of generality, suppose that $|X|=\mathbb{R}$. Let $h \colon \mathbb{R} \to X \times Y$ be a homeomorphism. It is easy to verify that $\pi_X \circ h \colon \mathbb{R} \to X$ is a homeomorphism, where $\pi_X \colon X \times Y \to X$ is the projection map.

Suppose to contradiction that $|Y|&gt;1$?

Pretty sure there's something you can do there but I'm gonna sleep on that for now.

Hope that helps.