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 $\alpha\in K$. ∎