The set of all rational points in the plane is a countable set

From Kolmogorov's Introductory Real Analysis. I am doing some self-study and would like some feedback on whether my proof is correct.

I am using that the set of rational numbers is countable as given, and I am invoking the following Theorem which is proved in the book.

Theorem 2. The union of a finite or countable number of countable sets $A_1, A_2, \ldots$ is itself countable.

Claim. The set of all rational points in the plane (points with rational coordinates) is countable.

Proof.
Since $\mathbb{Q}$ is a countable set, we can write $\mathbb{Q} = \{q_1, q_2, q_3, \ldots\}$. If we fix $q_1$ we can define the following set:$$Q_1 = \{(q_1, q)\;|\; q\in\mathbb{Q}\},$$which are all the rational points in the plane with $q_1$ in the $x$ position. We can create a one-to-one correspondence with $\mathbb{Q}$ by simply setting $p\leftrightarrow (q_1, p)$ for each $p\in\mathbb{Q}$, which showsthat $Q_1$ is countable. We can now define the following union which includes all rational points in the plane:$$\mathcal{Q} = \bigcup_{n=1}^\infty Q_n,$$which is a countable union of countable sets. By Theorem 2, $\mathcal{Q}$ is a countable set.■