Let $F$ be a field and assume that $[\overline F : F] = \infty$. Does it imply that for any $n \geq 1$, there is a field extension $K/F$ of degree $n$?
Notice that if $[\overline F : F] < \infty$ then $F$ is a real closed field so the answer is no (for $n > 2$). Moreover, it is not interesting to ask for extensions $\overline F/K$ such that $[\overline F : K] = [\overline K : K] = n$ because, by Artin-Schreier, this implies $n \leq 2$.
I know that there exist extensions $K/F$ with arbitrarily large degree (take $x_0 \in \overline F \setminus F$ then $K_0 = F(x_0)$ has finite degree over $F$, so we can find $x_1 \in \overline F \setminus K_0$, then $K_1=K_0(x_1)$ has finite degree over $F$ and so on).
Clearly, there are $F$-vector subspaces of $\overline F$ of dimension $n$ over $F$, but they might not be subfields.
I know that $L = \Bbb Q(\sqrt p \mid p \text{ prime}) / \Bbb Q$ has no sub-extension $K/\Bbb Q$ of degree $3$. But here $L$ is not the algebraic closure of $\Bbb Q$.