Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space on which independent random variables $X$ and $Y$ are defined. Suppose neither $X$ and $Y$ are almost surely constant. Must $\Omega$ be of the form $\Omega=\Omega_1\times\Omega_2$ for some sets $\Omega_1$ and $\Omega_2$?
Even in the simplest case of $X$ and $Y$ being $\text{Bernoulli}(1/2)$ I was unable to make them independent, at least using $\Omega=\mathbb{R}$.
No, $\Omega$ does not need to be a produce space.
On $\Omega = [0,1]$ (endowed with Lebesgue measure $\lambda$) consider the random variables
$$X := 1_{[0,1/4]} + 1_{[3/4,1]} \qquad Y := 1_{[1/2,1]}.$$
Clearly, $X$ and $Y$ are Bernoulli distributed with parameter $p=1/2$. Moreover,
$$\begin{align*} \mathbb{P}(X=1,Y=1) = \lambda \left( \left[ \frac{3}{4},1 \right] \right) &= \frac{1}{4} = \mathbb{P}(X=1) \mathbb{P}(Y=1) \\ \mathbb{P}(X=0, Y=1) = \lambda \left( \left[ \frac{1}{2}, \frac{3}{4} \right) \right) &= \frac{1}{4} = \mathbb{P}(X=0) \mathbb{P}(Y=1) \\ \mathbb{P}(X=1, Y=0) = \lambda \left( \left[ 0, \frac{1}{4} \right] \right) &= \frac{1}{4} = \mathbb{P}(X=1) \mathbb{P}(Y=0) \\ \mathbb{P}(X=0, Y=0) = \lambda \left( \left( \frac{1}{4}, \frac{1}{2} \right) \right) &= \frac{1}{4} = \mathbb{P}(X=0) \mathbb{P}(Y=0) \end{align*}$$ which shows that $X$ and $Y$ are independent.