Countability of set of all functions with domain $\mathbb{Z}$ and codomain $\mathbb{R}$ 
I’m not sure how to attempt this problem and would appreciate some guidance. I have the following, but I’m sure it’s incorrect:

 A: Because $\mathbb{N}$ and $\mathbb{Z}$ have the same cardinality, the space of functions from $\mathbb{Z}$ to $\mathbb{R}$ has the same size as the space of functions from $\mathbb{N}$ to $\mathbb{R}$ (this is the space of all real-valued-sequences).
Cantor's diagonality argument gives us that the space of all $\{0,1\}$-sequences is uncountable. Therefore the space of all real-valued-sequences is uncountable. Therefore the collection in question is uncountable.
A: Your proof would be complete if you actually show that the set has cardinality $\ge|\mathbb{R}|$. This is essentially obvious: for every $r\in\mathbb{R}$, define $\bar{r}\colon\mathbb{Z}\to\mathbb{R}$ to be the constant function $\bar{r}(x)=r$.
Then $r\mapsto\bar{r}$ is an injective map $\mathbb{R}\to \mathbb{R}^{\mathbb{Z}}$, proving that
$$
|\mathbb{R}|\le|\mathbb{R}^{\mathbb{Z}}|
$$
More generally, $|X|\le|X^Y|$ for every nonempty set $Y$, the argument is the same (where $X^Y$ denotes the set of all maps $Y\to X$).
Actually, also the converse inequality holds, in your case:
$$
|\mathbb{R}^{\mathbb{Z}}|=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\aleph_0}=
2^{\aleph_0}=|\mathbb{R}|
$$
