Generating the Borel $\sigma$-algebra on $C([0,1])$ We put $S=C([0,1])$ (the collection of continuous real functions on $[0,1]$), equipped with the metric $d(f,g)=\sup_{x\in[0,1]}|f(x)-g(x)|$, and let $\mathcal{B}(S)$ be the Borel $\sigma$-algebra on $S$. 
I'd appreciate some quidance on proving that $\mathcal{B}(S)$ is generated by the collection of sets of the form
$$
\{f: (f(t_1),\ldots, f(t_n)\in B_1\times\cdots \times B_n\},
$$
where $B_j\in \mathbb{B}(\mathcal(R))$, $n<\infty$ and $0\le t_1<t_2<
\cdots t_n\le 1$.
I know that $S$ is complete and separable, but need some help on how to proceed. 
 A: Consider the collection $\mathcal{C}$ consisting of all sets of the form $\{ f: (f(t_1),...,f(t_n)) \in B_1 \times \cdots \times B_n \} = \bigcap_{i=1}^n \pi_{t_i}^{-1} (B_i)$, where $n \in \mathbb{N}$, $t_i \in [0,1]$, and $B_i \in \mathcal{B}(\mathbb{R})$. 
First notice that $\mathcal{C} \subset \mathcal{B}(S)$. To show this note that if $t \in [0,1]$ and $(a,b) \subset \mathbb{R}$, then $\pi_t^{-1}((a,b))$ is open in $S$. In particular $\{ B \subset \mathbb{R} : \pi_t^{-1}(B) \in \mathcal{B}(S) \}$ is a sigma algebra containing all open intervals of $\mathbb{R}$, hence containing all Borel sets in $\mathbb{R}$. Thus $\pi_t^{-1}(B) \in \mathcal{B}(S)$ for all $t \in [0,1]$ and $B \in \mathcal{B}(\mathbb{R})$, and since elements of $\mathcal{C}$ are finite intersections of such sets, we have that $\mathcal{C} \subset \mathcal{B}(S)$.
To prove the reverse inclusion (i.e, $\mathcal{B}(S) \subset \sigma(\mathcal{C})$), note that since $S$ is separable, it suffices to prove that any open ball in $S$ is in $\sigma(\mathcal{C})$, because this would prove that any open set in $S$ is in $\sigma(\mathcal{C})$.
I claim that any open ball $B(f,\epsilon)$ in $S$ can be written as $$B(f,\epsilon) = \bigcup_{n \in \mathbb{N}} \; \;\bigcap_{t\in \mathbb{Q} \cap [0,1]} \pi_t^{-1}\big((f(t)-(1-2^{-n})\epsilon\;,\; f(t)+(1-2^{-n})\epsilon)\big)$$
This is true because if $f,g \in S$, and if there exists $n$ such that $|f(t)-g(t)|< (1-2^{-n})\epsilon$ for all $t \in \mathbb{Q} \cap [0,1]$, then by continuity of $f$ and $g$ and by density of $\mathbb{Q}$ in $[0,1]$, we know that $d(f,g)\leq (1-2^{-n})\epsilon<\epsilon$. And conversely, if $d(f,g)<\epsilon$, then there exists $n \in \mathbb{N}$ such that $d(f,g)<(1-2^{-n})\epsilon$, snd so $|f(t)-g(t)|<(1-2^{-n})\epsilon$ for all $t \in [0,1] \cap \mathbb{Q}$.
Therefore $B(f,\epsilon)$ can be written as a countable union of a countable intersection of elements of $\mathcal{C}$, and so $B(f, \epsilon) \in \sigma(\mathcal{C})$, as desired.
Additional Remarks: The map $\pi_t:S \to \mathbb{R}$ is the evaluation map at $t$, i.e, $\pi_t(f):=f(t)$.
Also, the [$1-2^{-n}$] can be replaced with any sequence $a_n \uparrow 1$, and also the collection $\mathcal{C}$ can be replaced with the smaller collection $\mathcal{C}_0:=\{\pi_t^{-1}(B) : t \in [0,1]$ and $B \in \mathcal{B}(\mathbb{R})\}$.
