Formally show that the set of continuous functions is not measurable

Here's a direct argument following tomasz's suggestion. Let's take as given your characterization of the measurable sets $\mathcal{B}(\mathbb{R}^\mathbb{R})$.

Let $A \in \mathcal{B}(\mathbb{R}^\mathbb{R})$ be arbitrary. We will show $A \ne C(\mathbb{R})$. If $A = \emptyset$ we are done. Otherwise, suppose $f \in A$. If $f$ is not continuous, we are done. Otherwise, suppose $f$ is continuous. We know $A$ is of the form $A = B \times \mathbb{R}^{\mathbb{R} \setminus J}$, where $B \subset \mathbb{R}^J$ and $J$ is at most countable. Since $\mathbb{R}$ is uncountable, $\mathbb{R} \setminus J$ is nonempty, so let $y \in \mathbb{R} \setminus J$. Define $g$ by $g(y) = f(y)+1$ and $g(x) = f(x)$ for $x \ne y$. Then $g \in A$, but $g$ is certainly not continuous.

Let consider a mapping $T:C(R)\to R^{Q}$ defined by $T(f)=(f(q))_{q \in Q}$, where $Q$ denotes a set of all rational numbers of $R$. It is obvious that $T$ is injective which implies that $card(C(R))\le card(R^Q)=(2^{\aleph_0})^{\aleph_0}= 2^{\aleph_0\times \aleph_0}=2^{\aleph_0}$, where $\aleph_0$ denotes the cardinality of all natural numbers.

If we assume that $C(R)$ is measurable with respect to product topology then there will be a countable parameter set $J \subset R$ and a non-empty Borel subset $A \subseteq R^J$ such that
$C(R)=A \times R^{R \setminus J}$. Notice that $$card(A \times R^{R \setminus J})\ge 1 \times card(R^{R \setminus J})=card(R^R)={2^{\aleph_0}}^{2^{\aleph_0}}=2^{2^{\aleph_0}}>2^{\aleph_0}$$ and we get the contradiction.