ii (source code)
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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.
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