Is "product" of Borel sigma algebras the Borel sigma algebra of the "product" of underlying topologies?

The Borel $\sigma$-algebra of the countable product of second countable topological spaces is the product of the Borel $\sigma$-algebras. it is always true that the Borel $\sigma$-algebra of the topological product space is at least as large as the product of the $\sigma$-algebras. Proofs of theis can be found for example in Kallenberg's book.

The Borel $\sigma$-algebra of the uncountable product of nontrivial (at least two points) Hasudorff spaces is always larger than the product of the $\sigma$-algebras. To see this, note that every point is closed in the product topology and therefore a Borel set. But by the construction of the product $\sigma$-algebra, a set can depend only on countably many coordinates. More precisely, there is a general result that if $A\in\sigma(\mathcal{F})$ then there is a countable family $\mathcal{C}\subseteq\mathcal{F}$ such that $A\in\mathcal{C}$. To prove this, just check that the sets generated by a countable subfamily of $\mathcal{F}$ give you a $\sigma$-algebra containing $\mathcal{F}$. In particular, every set in the product $\sigma$-algebra is generated by countably many measurable rectangles.

It is asserted (with a short proof) in Billingsley's Convergence of Probability measures (second edition) on page 244, that this holds if the underlying spaces are separable. (He only considers the product of two spaces).