Let $X$ be a countable set and $Y\subset X$. If $X$ or $Y$ is finite, we are actually done quickly. Therefore, we only consider the case where $X$ is countably infinite and the subset is not finite. Suppose $X=\{x_0,x_1,x_2,\ldots\}$. Then there is a smallest $i$, say $i_0$, for which $x_i\in Y$ and we denote $x_{i_0}$ as $y_0$. Furthermore, there is a smallest $i$, say $i_1$, for which $x_{i_1}\in Y\backslash \{y_0\}$ and we denote $x_{i_1}$ as $y_1$. If $y_0,y_1, \ldots y_n$ have already been defined, we take $y_{n+1}$ to be the element $x_{i_{n+1}}$, where $i_{n+1}$ is the smallest index $i$ for which $x_i\in Y\backslash \{y_0,y_1,\ldots y_n\}$. Note that every element of $Y$ appears in the sequence $y_0,y_1,\ldots$ and that this sequence does not end, as $Y$ is not assumed to be finite. Thus, $Y$ is countably infinite. Therefore, $Y$ is a countable set in all cases.

We prove the countability of $X\cup Y$ only for two sets that are countably infinite, so if $X\cong \mathbb{N}$ and $Y\cong \mathbb{N}$. Because $\mathbb{N}\cong \{n\in \mathbb{N}\mid n\text{ is even}\}$ and $\mathbb{N}\cong \{n\in \mathbb{N}\mid n\text{ is odd}\}$, we can create a bijective function from $X\cup Y$ to $\mathbb{N}$ by mapping $X$ bijectively to $\mathbb{N}\mid n\text{ is even}\}$ and $Y$ bijectively to $\mathbb{N}\mid n\text{ is odd}\}$. In other words, $X\cup Y$ is countable.

We prove the countability of $X\times Y$ only for two sets that are countably infinite, so if $X\cong \mathbb{N}$ and $Y\cong \mathbb{N}$. Then, there is a bijection from $X\times Y$ to $\mathbb{N}\times \mathbb{N}$. Thus, $X\times Y\cong\mathbb{N}\times \mathbb{N}$. Because $\mathbb{N}\times \mathbb{N}\cong (\mathbb{N}$, we have $X\times Y\cong\mathbb{N}$. In other words, $X\times Y$ is countable.

We first show that the set $\mathbb{Q}^{+}$ of all positive rational numbers is countably infinite. These positive rational numbers can be considered as $p/q$ with $p$ and $q$ positive natural numbers and $p/q$ an irreducible fraction, i.e., $\mathrm{gcd}(p,q)=1$. We can now define the function $f:\;\mathbb{Q}\rightarrow \mathbb{Z}\times \mathbb{Z}\backslash\{0\}$ by $f(p/q)=(p,q)$. Then $f$ is an injective function. Because $\mathbb{N}\times \mathbb{N}$ is countably infinite and $f(\mathbb{Q}^{+}$ is a infinite subset of this, it follows that $f(\mathbb{Q}^{+})$ is countably infinite. But then $\mathbb{Q}^{+}$ is also countably infinite.

The set $\mathbb{Q}^{-}$ of all negative rational numbers is evidently equipotent to $\mathbb{Q}^{+}$ and thus also countably infinite. Therefore, $\mathbb{Q}=\mathbb{Q}^{-}\cup\{0\}\cup \mathbb{Q}^{+}$, as union of countable sets, is also countable. Because the set of rational numbers is not finite, we know that this set is countably infinite.