I've been given some statements to prove using only the following facts.
The set of all integers is countably infinite.
Let $X$ and $Y$ be sets and suppose $f:X\rightarrow Y$ is surjective. If $X$ is countable, then $Y$ is also countable.
The union of a countably infinite family of sets is countable.
I'd like your opinion on my proofs. I feel like the first could be simpler.
Exercise 1. The set $\mathbb{Z}\times\mathbb{Z}$ is countable.
Proof. Let $A_n=\{n\}\times\mathbb{Z}$ for each $n\in\mathbb{Z}$. Define a mapping $f:\mathbb{Z}\rightarrow A_n$ by $f(x)=(n,x)$ for all $x\in\mathbb{Z}$. To see that $f$ is surjective, consider $(n,p)\in A_n$. Since $p\in\mathbb{Z}$, $f(p)=(n,p)$. Thus, $f$ is surjective. By (2), since (1) $\mathbb{Z}$ is countable, $A_n$ is countable.
Let $S=\{A_n:n\in\mathbb{Z}\}$. Notice that $S$ is infinite since it is indexed over $\mathbb{Z}$. Define $g:\mathbb{Z}\rightarrow S$ by $g(x)=A_x$ for all $x\in\mathbb{Z}$. To see that $g$ is surjective, consider $A_q\in S$. Since $q\in\mathbb{Z}$, $g(q)=A_q$. Thus, $g$ is surjective. By (2), $S$ is countable. Thus, $S$ is a countably infinite family of countable sets.
Let $U=\bigcup S$. Then, by (3), $U$ is countable. To see that $U=\mathbb{Z}\times\mathbb{Z}$, we will show that $U$ and $\mathbb{Z}\times\mathbb{Z}$ are subsets of each other.
Let $x\in U$. Then $x\in A_n$ for some $n\in\mathbb{Z}$. Hence, $x=(n,p)$ for some $p\in\mathbb{Z}$. Therefore, $x=(n,p)\in\mathbb{Z}\times\mathbb{Z}$, and it follows that $U\subseteq\mathbb{Z}\times\mathbb{Z}$.
Let $x\in\mathbb{Z}\times\mathbb{Z}$. Then $x=(p,q)$ for some $p,q\in\mathbb{Z}$. Thus, $x\in A_p$, which implies that $x\in U$. Therefore, $\mathbb{Z}\times\mathbb{Z}\subseteq U$, and it follows that $U=\mathbb{Z}\times\mathbb{Z}$.
Therefore, $\mathbb{Z}\times\mathbb{Z}$ is countable.
Exercise 2. Every infinite subset of a countable set is countable.
Proof. Let $X$ be a countable set and let $Y$ be an infinite subset of $X$. Then $X$ is countably infinite and there exists a surjection $f:\mathbb{N}\rightarrow X$. Since $Y\subseteq X$ and $Y$ is nonempty, there exists a surjection $g:X\rightarrow Y$. Thus, $gf:\mathbb{N}\rightarrow Y$ is a surjection. Therefore, by (2), $Y$ is countable.