Example of when $\mathcal{B}(X\times Y) \neq \mathcal{B}(X) \times\mathcal{B}(Y)$ but $|X|,|Y| \leq |\mathbb{R}|$ I am interested in knowing examples of when $\mathcal{B}(X\times Y) \neq \mathcal{B}(X) \times\mathcal{B}(Y)$. By allowing $|X|,|Y|$ to be large we can provide a trivial counterexample, as in the one given at https://mathoverflow.net/questions/39882/product-of-borel-sigma-algebras. I know that if $X,Y$ are separable metrizable then the the product equality holds. Do we have any counterexamples where $|X|,|Y| \leq |\mathbb{R}|$?
In particular I am interested in knowing whether $X=Y=C(\mathbb{R})$ (equipped with the subspace topology inherited from $\mathbb{R}^{\mathbb{R}}$) provides the necessary counterexample.
 A: The article by Arnold Miller that was suggested by Martin in the linked answer shows that it is consistent to have $\mathcal{P}(2^\omega) \otimes \mathcal{P}(2^\omega) \neq \mathcal{P}(2^\omega \times 2^\omega)$, specifically this is Theorem 56. A universal analytic set is an example of a "set universal for the Borel sets" that is needed in that proof, and these are well-known to exist (you can find this in Jech's Set Theory, for example).
Miller also says that Kunen proved, in his thesis, that if you add $\aleph_2$ Cohen reals to a model of set theory satisfying the generalized continuum hypothesis, then any well-ordering of $\omega_2$ is a subset of $\omega_2 \times \omega_2$ not contained in $\mathcal{P}(\omega_2) \otimes \mathcal{P}(\omega_2)$. Adding $\aleph_2$ Cohen reals to a model with GCH makes $2^\omega = \aleph_2$, so this is another way of showing this.

I now address your more specific question about $C(\mathbb{R})$ with the pointwise $\sigma$-algebra. This is not a counterexample because, as I shall show, it is isomorphic to the Borel $\sigma$-algebra of a separable metric space (this follows from some general theorems known to experts, but I'll give a proof anyway).

The $\sigma$-algebra you suggest for $C(\mathbb{R})$ is generated by the sets 
$$
G_{x,U} = \{ f : \mathbb{R} \rightarrow \mathbb{R} \mid f(x) \in U \},
$$
where $x \in \mathbb{R}$ and $U \subseteq \mathbb{R}$ is an open interval with rational endpoints. We will call this $\Sigma_{\mathbb{R}}$. We can also limit $x$ to range over $\mathbb{Q}$, and in this way obtain a countably generated $\sigma$-algebra $\Sigma_{\mathbb{Q}}$. As a countably generated $\sigma$-algebra, $\Sigma_{\mathbb{Q}}$ is the Borel $\sigma$-algebra of a separable metric space (alternatively, we can consider ourselves to have measurably embedded $C(\mathbb{R})$ in $\mathbb{R}^{\mathbb{Q}}$, a Polish space). Therefore we are done if we can show $\Sigma_{\mathbb{R}} = \Sigma_{\mathbb{Q}}$. Clearly $\Sigma_{\mathbb{Q}} \subseteq \Sigma_{\mathbb{R}}$.
To show the opposite inclusion, we show that each $G_{x,U}$ for $x \in \mathbb{R}$ is contained in $\Sigma_{\mathbb{R}}$. We know that there exists a sequence $(x_i)$ of rationals such that $x_i \to x$, and then all we need is the continuity of each $f \in C(\mathbb{R})$:
$$
G_{x,U} = \{ f \in C(\mathbb{R}) \mid f(x) \in U \} = \{ f \in C(\mathbb{R}) \mid \exists N \in \mathbb{N} . \forall i \geq N . f(x_i) \in U \} = \bigcup_{N \in \mathbb{N}} \bigcap_{i \geq N} G_{x_i,U},
$$
which, by the rationality of each $x_i$, is an element of $\Sigma_{\mathbb{Q}}$.
