Some background. I was looking for a product of two Borel $σ$-algebras which is not Borel. I know that the two topological spaces mustn't be second-countable. I then googled an example of such a space and found the long line on Wikipedia.
Now the problem. The large plane is the topological product of two long lines. The disjoint union of uncountably many unit squares, let's say
$$A:=\bigcup_{α∈ω_1} \bigl] (α,0),(α,1) \bigr[^2,$$
seems (until I'm proved wrong...) a reasonable bet for an open set which is not in the product of the Borel $σ$-algebra of the long line with itself. However, I'm still not able to prove it. Should I give up? Is my conjecture true?
I don’t know if the product of two long lines works, but here is a somewhat simpler and more straightforward counterexample.
Let $X$ be any set whose cardinality is greater than that of the continuum (for example, the power set of $\mathbb R$). Endow it with the discrete topology. Let $D\equiv\{(x,x)\mid x\in X\}$ be the diagonal of $X\times X$. Then, $D$ is clopen (as is any subset of $X\times X$) in the product topology, so it lies in $\mathscr B_{X\times X}$ (which is, of course, the discrete $\sigma$-algebra on $X\times X$).
However, $D\notin\mathscr B_X\otimes\mathscr B_X$. If $D$ were in the product of the Borel $\sigma$-algebras, then there would exist a $\sigma$-subalgebra $\mathscr M\subseteq\mathscr B_X\otimes\mathscr B_X$ that is generated by a countable class of measurable rectangles and $D\in\mathscr M$. It follows that there exists a countably generated $\sigma$-subalgebra $\mathscr N\subseteq\mathscr B_X$ with $D\in\mathscr N\otimes\mathscr N$. Then, since $\{x\}$ is a coordinate projection of $D$ for every $x\in X$, one would have that $\{x\}\in\mathscr N$. This is impossible, given that $\mathscr N$ is countably generated, so that its cardinality cannot be greater than that of $\mathbb R$.
For more details, see Folland (1999), in particular: Exercise 7.29 (page 231), Exercise 1.5 (page 24), and Proposition 1.23 (page 40).