My question is: How does choosing $N$ sufficiently large effect this proof?
Prove that $\phi : \prod_{i=1}^{\infty} (A_i,T_i)$ onto $[0,1]$ is continuous.
Let each $(A_i,T_i) = (\{0,2\}, T_{discrete})$ and $\phi (<a_1, a_2, ...>) = \sum^{\infty}_{i=1} \frac{a_i}{2^{i+1}}$.
Then to prove it is continuous show that if $U$ is the open interval $$\sum^{\infty}_{i=1} \frac{a_i}{2^{i+1}} \in (\sum^{\infty}_{i=1} \frac{a_i}{2^{i+1}} - \epsilon, \sum^{\infty}_{i=1} \frac{a_i}{2^{i+1}} + \epsilon)$$ for any $\epsilon > 0$ then there exists an open set $(<a_1, a_2, ...>) \in W$ such that $\phi (W) \subseteq U$.
Choose $N$ sufficiently large such that $\sum^{\infty}_{i=N} \frac{a_i}{2^{i+1}} \lt \epsilon$ and put $W = \{a_1\} \times \{a_2\} \times ... \times \{a_N\} \times A_{N+1} \times ...$
Then $W$ is open in $\prod_{i=1}^{\infty} (A_i,T_i)$, $<a_1, a_2, ...> \in W$, and $\phi (W) \subseteq U$, as required.