If $X$ and $Y$ are countable, then $X \times Y$ is countable. (Analysis 1 by Tao) Exercise 8.1.8 Use Corollary 8.1.13 to prove Corollary 8.1.14.
Corollary 8.1.13. The set $\mathbb{N} \times \mathbb{N}$ is countable.
Corollary 8.1.14. If $X$ and $Y$ are countable, then $X \times Y$ is countable.

For the proof of Corollary 8.1.13, the book shows that $\mathbb{N} \times \mathbb{N} = A \cup B$, where $A = \{(n,m) \in \mathbb{N} \times \mathbb{N} : 0 \le m \le n\}$ and $B = \{(m,n) \in \mathbb{N} \times \mathbb{N} : 0 \le n \le m\}$.
I also know that $X$ and $Y$ are countable, there exists bijections from $\mathbb{N}$ to $X$ and $\mathbb{N} $ to $Y$. To finish the proof, I need to find the bijection from $\mathbb{N} \times \mathbb{N}$ to $X \times Y$, but I do not know how to get this.
Any help would be appreciated.
Denote by $f:\Bbb N \to X$ , $g:\Bbb N \to Y$ the bijections from $\Bbb N$ to $X,Y$.

Then $h : \Bbb N \times \Bbb N \to X \times Y$ which is defined by $h(n,m)= (f(n),g(m))$ is the bijection you are looking for.
As this was downvoted, I will provide another argument. In the following argument I will allow countability to include finite and countably infinite.

Before starting, a useful lemma is the following:

If $f:X\rightarrow Y$ is injective and $Y$ countable then $X$ is countable.

Proof:

Suppose $Y$ is finite. Observe, $f:X\rightarrow f(X)$ is bijective and since $f(X)$ is finite, so is $X$. Hence $X$ is countable. Suppose now that $Y$ is countably infinite. Then $f(X)\subseteq Y$ is finite or countably infinite (countable). In either case, we admit a bijection from $X$ to $f(X)$, in which case $X$ is either finite or, if $f(X)$ is countably infinite, we obtain a bijection $g:f(X)\rightarrow \mathbb{N}$ and as a result, considering $f:X\rightarrow f(X)$, we obtain a bijection, $g\circ f: X\rightarrow \mathbb{N}$.

Now, with regards to your problem,

Suppose $X$ and $Y$ are both countable.

Case 1: Suppose $X$ is finite and $Y$ is countably infinite.

In this case, one may define an injection, $f: X \rightarrow \mathbb{N}$ (Exercise: Find such a injection) and a bijection, $g:Y\rightarrow \mathbb{N}$. Therefore, define a function $h: X\times Y \rightarrow \mathbb{N} \times \mathbb{N}$  by $h(x,y)$ $=$ ($f(x),g(y))$. It is straightforward to check that this function is injective.

Case 2: Suppose $X,Y$ are both countably infinite. In which case, we are able to obtain bijective maps, from $X$ to $\mathbb{N}$ and $Y$ to $\mathbb{N}$. Therefore, repeating what we did in case 1 gives us the desired conclusion.

Other cases follow similarly.