Product of Borel sigma algebras If $X$ and $Y$ are separable metric spaces, then the Borel $\sigma$-algebra $B(X \times  Y)$ of the product is the $\sigma$-algebra generated by $B(X)\times B(Y)$.  I am embarrassed to admit that I don't know the answers to: 
Question 1. What is a counterexample when $X$ and $Y$ are non-separable?
Question 2. If $X$ is an uncountable discrete metric space, does
$B(X) \times B(X)$ generate the Borel $\sigma$-algebra on $X \times X$?
Question  3. If $X$ and $Y$ are metric spaces, with $X$ separable, does
$B(X) \times B(Y)$ generate the Borel $\sigma$-algebra on $X \times Y$?
 A: The answer to question 3 is yes. 
At least according to Lemma 6.4.2 of the second volume of Bogachev's book "Measure Theory".
He requires both spaces to be Hausdorff and one of them to have a countable base. 
They need not be metric spaces. 
A: Q1.  Discrete spaces with cardinal > c ... then the diagonal is a Borel set, but not in the product sigma-algebra.
This also answers Q2 (no)
but not Q3.
A: To close a gap: From the answer of Gerald Edgar, we know that the answer to the second question is no if the spaces involved have cardinality larger than $\mathfrak{c}$. This leaves open what happens when they do have cardinality $\mathfrak{c}$. The answer is yes under the continuum hypothesis, and in general it holds that $2^{\omega_1}\otimes 2^{\omega_1}=2^{\omega_1\times\omega_1}$. This was shown in
B. V. Rao, On discrete Borel spaces and projective sets
Bull. Amer. Math. Soc. Volume 75, Number 3 (1969), 614-617.
In Bogachev's remarkable book, it can be found as Proposition 3.10.2.
A: This is should probably rather be a comment to Michael Greinecker's answer, but I do not have the necessary privileges.
Michael Greinecker's answer leaves open what happens with a continuum-sized discrete space when one does not assume the continuum hypothesis.
Arnold W. Miller showed in section 4 of On the length of Borel hierarchies that it is consistent relative ZFC that no universal analytic set $U \subset [0,1] \times [0,1]$ belongs to the product $\sigma$-algebra $\mathcal{P}[0,1] \otimes \mathcal{P}[0,1]$. Combined with Rao's result mentioned by Michael Greinecker, this shows that $2^{\mathfrak{c \times c}} = 2^{\mathfrak{c}} \otimes 2^\mathfrak{c}$ is independent of ZFC.
See my answer to Universally measurable sets of $\mathbb{R}^2$ on math.stackexchange.com for related results and more details and references.
