If you know that $\mathbb Q$ is dense in $\mathbb R$, then a proof using countable choice is straightforward. Construct a nested sequence of open intervals containing the sought $x\in \mathbb{R}\setminus \mathbb{Q}$ whose diameters shrink to zero, say $U_n = (x-1/n,x+1/n)$.
Choose a rational number $q_n \in U_n$ for each open interval. Since $|q_n - x| \lt (1/n)$, necessarily $\lim_{n\to \infty} q_n = x$.
A more "constructive" proof, avoiding countable choice, is outlined in the Comment by @Cia, but it requires some machinery of decimal expansions that will have to be separately justified. Alternatively we can choose a "least" rational number in the sense of a lexicographic ordering (least denominator, followed by least numerator) in each $U_n$ as another approach to avoiding countable choice.