Set of continuous function is not measurable 
I would like to prove that $\mathcal{C}([0,1])\notin \mathcal{B}^{[0,1]}.$

For all $I\subset[0,1]$, denote $\mathcal{B}_E(I)$ the smallest $\sigma-$algebra of $E^[0,1]$ such that all $\pi^t,\; t\in I$ are measurable. 
Denote $D$ the set of countable sets of $[0,1].$
I know that $\mathcal{B}^{[0,1]}=\cup_{I\in D}\mathcal{B}_E(I).$
I imagine that we argue by contradiction and we can find a function such that $\pi^t$ are not measurable, but not sure I understand correctly the problem ?
 A: Roughly speaking, $\mathcal B^{[0,1]}$ consists of all sets that depend only on countably many coordinates. This is very informal, so we need to be more precise.
Say that a set $A\subseteq \mathbb R^{[0,1]}$ satisfies property $(\star)$ if there exists a countable set $F=F(A)$ such that, for all $f\in A$ and $g\in\mathbb R^{[0,1]}$, if $f|_F=g|_F$ then $g\in A$. Let $\mathcal A$ be set of all $A\subseteq\mathbb R^{[0,1]}$ that satisfy property $(\star)$. Note that all sets of the form $\{f:f(x)\in B\}$, with $x\in[0,1]$ and $B\subseteq\mathbb R$ Borel measurable, satisfy property $(\star)$. It will follow that $\mathcal B^{[0,1]}\subseteq\mathcal A$ if we can show $\mathcal A$ is a $\sigma$-field. This will complete the proof since clearly $C([0,1])$ does not satisfy property $(\star)$. (Indeed, for any $f\in C([0,1])$ and any $x\in[0,1]$ there exists $g\in\mathbb R^{[0,1]}$ such that $f(y)=g(y)$ for all $y\neq x$, but $g$ is not continuous at $x$.)
First, we need $\emptyset\in\mathcal A$. This is trivial; take $F(\emptyset)=\emptyset$, or any countable set.
Next, suppose $A\in\mathcal A$. We need to show $A^c\in\mathcal A$. I claim $A^c$ satisfies property $(\star)$ with $F(A^c)=F(A)$. Indeed, suppose $f\in A^c$ and $g|_{F(A)}=f|_{F(A)}$. Assume for a contradiction that $g\notin A^c$. Then $g\in A$, so since $A$ satisfies property $(\star)$, it follows that $f\in A$, a contradiction. Thus $g\in A^c$, which implies $A^c$ satisfies property $(\star)$, that is, $A^c\in\mathcal A$.
Finally, suppose $A_i\in\mathcal A$ for $i\in\mathbb N$. We need to show $A:=\bigcup_{i\in\mathbb N}A_i\in\mathcal A$. I claim $A$ satisfies property $(\star)$ with $F(A)=F_0:=\bigcup_{i\in\mathbb N}F(A_i)$. (Observe that $F_0$ is indeed countable.) Indeed, suppose $f\in A$ and $g|_{F_0}=f|_{F_0}$. Since $f\in A$, there exists $i_0\in\mathbb N$ such that $f\in A_{i_0}$. Since $g|_{F_0}=f|_{F_0}$ and $F_{i_0}\subset F_0$, it follows that $g|_{F_{i_0}}=f|_{F_{i_0}}$, which by property $(\star)$ implies $g\in A_{i_0}\subset A$. Hence $A$ satisfies property $(\star)$, that is, $A\in\mathcal A$, so $\mathcal A$ is a $\sigma$-field and the proof is complete.
