= Solution
An <algebraic closure> of a field $K$ is an <algebraically closed field> that is an algebraic extension of $K$.
By definition, every element of
$$
\overline{\mathbb Q}
=\{z\in\mathbb C:z\text{ is algebraic over }\mathbb Q\}
$$
is 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 $\mathbb Q$.
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