$F$ is algebraically closed $\iff$ $\nexists$ $K$ s.t. $F \leq K$, $K \neq F$ and $[K:F] < \infty$ 
$F$ is algebraically closed $\iff$ $\nexists$ $K$ s.t. $F \leq K$, $K \neq F$ and $[K:F] < \infty$.

Proof:
($\implies$) assume by way of contradiction that $F$ is an algebraically closed field and there does exist such a field $K$, say $[K:F] = n$.
Let $a_1,....a_n$ be a basis for $K$ over $F$.
Okay, now I'm lost, haha
$(\impliedby)$ 
I need help with this direction also.
Thanks all! You're the best!

Depends on your definition for algebraically closed fields, but let me give it a try. Choose an element $a \in K$ which is not in $F$ and consider the corresponding minimal polynomial over $K$. Since $a \not\in F$, the minimal polynomial does not split over $F$ which is a contradiction to $F$ being algebraically closed.
For the other direction you can assume that $F$ is not algebraically closed and thus construct a finite extension by adjoining a root.
