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.
 A: 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. 
A: Suppose $X$ and $Y$ each have more than one point and $\mathbb R \simeq X\times Y$. 
Using the coordinate projections, we see that $X$ and $Y$ are both connected. 
Let $a,b\in X$ and $c,d\in Y$ be distinct points.  
Consider the points $p=(a,c)$ and $q=(a,d)$. 
Let
$$C_1=\{a\}\times Y$$
$$C_2=\big(\{b\}\times Y\big)\cup \big(X\times \{c\}\big)\cup \big(X\times \{d\}\big).$$
Both of these sets are connected, and $C_1\cap C_2=\{p,q\}$. 
But can two different points in $\mathbb R$ be joined by connected sets like this?
A: If $X$ and $Y$ are connected topological spaces, each containing at least two points, then the product space $X\times Y$ has no cut point.
Proof. Consider any point $(a,b)\in X\times Y;$ I have to show that $X\times Y\setminus\{(a,b)\}$ is connected.
Choose $x_0\in X\setminus\{a\}$ and $y_0\in Y\setminus\{b\}.$ Now consider any point $(x,y)\ne(a,b);$ I will show that $(x,y)$ and $(x_0,y_0)$ are in the same component of $X\times Y\setminus\{(a,b)\}.$
Case I. If $x\ne a$ then $(\{x\}\times Y)\cup(X\times\{y_0\})$ is a connected subset of $X\times Y\setminus\{(a,b)\}$ containing $(x,y)$ and $(x_0,y_0).$
Case II. If $y\ne b$ then $(X\times\{y\})\cup(\{x_0\}\times Y)$ is a connected subset of $X\times Y\setminus\{(a,b)\}$ containing $(x,y)$ and $(x_0,y_0).$
A: 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 $|Y|>1$?
Pretty sure there's something you can do there but I'm gonna sleep on that for now.
Hope that helps.
