Algebraic closures of subextensions are all the same?

Question

Let $L|K$ be a finite extension and $K \subseteq E \subseteq L$ and $K \subseteq F \subseteq L$ two subextensions.

Then $L|E$ and $L|F$ are finite and $\bar{L}|E$ and $\bar{L}|F$ are algebraic extensions with $\bar{L}$ being the algebraic closure of $L|K$. From this then follows $\bar{E}=\bar{F}=\bar{L}$. (Why is that, btw?)

Is this only true, when all the extensions are finite?

Answer 1

Finite implies algebraic, hence $L/E$ and $L/F$ are algebraic, from which the statement follows. So you $L/K$ doesn't need to be finite, just algebraic.