The real line is not homeomorphic to any non-trivial product space

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.

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 $(x_0,y_0).$