Let E be an extension of k which is algebraically closed and let A be the intersection of all the subextensions of E which are algebraically closed. I guess that A is actually the algebraic closure of k but I'm not quite sure. Is it correct?
Edit: I should apologize for the misunderstanding caused by the inexact description of my question. What I actually want to ask is how we can prove A is an algebraic closure of k, if we haven't proved the existence of an algebraic closure of k? In particular, I have just seen a proof which shows that A, as the intersection of all algebraically closed subextensions of E, is an algebraically closed field. So I am hoping for a proof that A is algebraic over k, given that we haven't known anything about the algebraic closure. Thanks!
The algebraic closure of $k$ inside $E$ (i.e. elements of $E$ which are algebraic over $k$), which is also an algebraic closure of $k$ (in the abstract sense) is an algebraically closed subextension of $E$ and is contained in every algebraically closed subextension of $E$, so the answer is yes.