 Solution

An algebraic closure of a field $K$ is an algebraically closed field that is an algebraic extension of $K$.

By definition, every element ofis algebraic over $\mathbb{Q}$, and the question allows us to use that this set is a field. It remains to prove that it is algebraically closed.

Let $f\in \overline{\mathbb{Q}}[X]$ be nonconstant. Its finitely many coefficients generate a finite extension $E/\mathbb{Q}$. By the fundamental theorem of algebra, $f$ has a root $z\in \mathbb{C}$. Since $z$ is a root of a polynomial over $E$, it is algebraic over $E$; since $E/\mathbb{Q}$ is algebraic, transitivity of algebraic extensions makes $z$ algebraic over $\mathbb{Q}$. Thus $z\in \overline{\mathbb{Q}}$. Repeating after division by $X-z$ shows that $f$ splits there, so $\overline{\mathbb{Q}}$ is algebraically closed and is an algebraic closure of $\mathbb{Q}$.