$C$ is the set of all infinite binary sequences - that is, an element of $C$ looks like $(b_1, b_2, b_3, . . .)$, where each $b_i$ is either $0$ or $1$. So for instance $(0, 1, 0, 1, 0, 1, . . . )$ is an element of $C$, and in this case $b_1=0, b_2=1, b_3=0, . . .$
Re: your first question, you're trying a bit too hard - there's a simpler way. Suppose I have a set $X\subseteq \mathbb{N}$, and you're trying to figure out what it is. The "brute force" approache would be to ask a series of questions:
"Is $1\in X$?"
"Is $2\in X$?"
"Is $3\in X$?"
Etc.
Each of these questions has a yes/no answer, and $X$ is determined by what my answers to each of these questions is. Do you see a way to use this idea to associate to $X$ a sequence of $0$s and $1$s?
For your second question, here's a hint: think about decimal (or other base!) expansions. This won't help you build a bijection right off, but it will help you build an injection from $C$ to $[0, 1]$, and a surjection from $C$ to $[0, 1]$, at which point you'll be able to use the Cantor-Bernstein theorem to show that there is a bijection (this theorem, I think, is the closest thing to an answer you'll get for your sub-question about whether there's a "general way" to show a bijection exists).
Yes this is fine. Although just to prove that $\Bbb{N^Z}$ is uncountable you only need to argue that $\aleph_0^{\aleph_0}\geq 2^{\aleph_0}$, which is trivial, and then that $2^{\aleph_0}$ is already uncountable.
Best Answer
Take the 1-to-1 function $f$ of the open interval $]0,1[$ in $\mathbb R$ defined by
$f(x)=$ representation of x in the numerical system of base $2$
This is enough.