 ii

An algebraically closed field is a field in which every nonconstant polynomial has a root, equivalently every such polynomial splits. An algebraic closure of $K$ is an algebraic extension of $K$ which is algebraically closed.

For the displayed subset $\overline{K}\subseteq \mathbb{C}$, if $\alpha ,\beta$ are algebraic over $K$, then $K(\alpha ,\beta )/K$ is finite. Sums, differences, products and nonzero inverses of its elements are algebraic over $K$, so $\overline{K}$ is a subfield. If a nonconstant polynomial has coefficients in $\overline{K}$, those finitely many coefficients generate a finite extension $F/K$. It has a root $z\in \mathbb{C}$ by algebraic closure of $\mathbb{C}$, and $z$ is algebraic over $F$. Part (i) makes it algebraic over $K$, so $z\in \overline{K}$. Hence $\overline{K}$ is algebraically closed and is an algebraic closure of $K$.

Every $z\in \overline{K}$ lies in the finite subextension $K(z)$, and every element of a finite subextension is algebraic over $K$. Therefore $\overline{K}$ is exactly the union of all finite subextensions inside $\mathbb{C}$. The two examples follow.