Show that infinite Cartesian product is uncountable let $A_i=\{0,1\} \forall i \in \mathbb Z^+$ and $A^{\omega}= \prod _\limits{i \in \mathbb Z^+}A_i$ thus $A^\omega=\{(a_i)|a_i=1,2 \forall i\in \mathbb Z^+\}$
now here is my problem

let $(\underline x_n)_{n \in \mathbb Z^+}$ where $\underline x_n=(x_{nk})_{k \in \mathbb Z^+}$ be a sequence in $A^\omega$. Define $\underline y=(y_n)_{n \in \mathbb Z^+}$ by $y_n=1-x_{nn}$ for all $n \in \mathbb Z^+$
a) Show that $\underline y \in A^\omega $ and $\underline y \neq \underline x_n$ for all $n\in \mathbb Z^+$
b) $\mathbf{using}$ $\mathbf{a)}$ show that $A^\omega$ is not countable

I did the a) part. And I can prove the part b) $A^\omega$ is uncountable by using diagonal argument. but here i should use the part a). my attempt was
first, I suppose $ A^\omega$ is countable.
then there exists a bijection from $f:\mathbb Z^+\to A^\omega$.
then I can list out all elements in  $A^\omega$.
then since I should use a) I have to find a connection between this list and $\underline x_n$s. I am stuck here.

Hint: If $A^{\omega}$ was countable, you could choose $(\underline{x}_n)_{n \in \mathbb{Z^+}}$ to take every value of $A^{\omega}$ (Because there is a bijection between $A^{\omega}$ and $\mathbb{Z}^+$). Now, why is that impossible from (a)? (Notice that this is exactly the diagonal argument)
