Proposition 1   If $K$ is a field, the following are equivalent:

(1)

$K$ is algebraically closed, i.e. every nonconstant polynomial $f$ in $K[x]$ has a root in $K$

(2)

Every nonconstant polynomial $f$ in $K[x]$ splits completely over $K$

(3)

If $L|K$ is an algebraic extension then $L = K$

Proof. If (1) is true then we can prove by induction on degree of $f$ that every nonconstant polynomial $f$ splits completely over $K$ Conversely, (2)$\Rightarrow$ (1) is trivial.
(2)$\Rightarrow$ (3) If $L|K$ is algebraic and $\alpha\in L$ then $\alpha$ is a root of a polynomial $f\in K[x]$ By (2) $f$ splits over $K$ which implies that $\alpha\in K$ It follows that $L=K$
(3)$\Rightarrow$ (1) Let $f\in K[x]$ and $\alpha$ a root of $f$ (in some extension of $K$ . Then $K(\alpha)$ is an algebraic extension of $K$ hence $\alpha\in K$