It depends on how you're topologizing the product space--but I'll assume you mean you're giving it the product topology. Then, let be Borel in . Then, by definition there exists countably many open sets such that, say for this case, . But, by definition of the product topology each may be written as where are open in respectively. But, for any second countable topological space (and thus in particular any separable metric space--since second countability and separability are equivalent for metric spaces) every open cover of a subset has a countable subcover (this is Lindelof's theorem--a proof can be found here at problem 26). Thus, we may assume that . The result then just falls out since