STATEMENT: This is a portion from Lang's proof of theorem 2.8 in chapter V section 2.
If $E$ is algebraically closed, and $L$ is algebraic over $\sigma k$, then $\sigma E$ is algebraically closed and $L$ is algebraic over $\sigma E$, hence $L=\sigma E$. Where $\sigma: k\rightarrow L$ is an embedding.
QUESTION: I wanted to ask if algebraically closed fields are necessarily minimal. That is if $F$ is an algebraically closed fields, does there exists a proper subfield that is also algebraically closed?
No. The algebraic closure of the rationals is countable, and is contained in $\Bbb C$ (the algebraic closure of the reals), which is uncountable. Any algebraically closed field does contain a minimal algebraically closed subfield: the algebraic closure of its prime subfield.
In response to your second question, note that any algebraic extension of an algebraically closed field is trivial (and since $L/\sigma E$ is algebraic, $L = \sigma E$); this is because, given $x \in L$, $x$ is algebraic over $\sigma E$, hence contained in $\sigma E$ by algebraic closure. The example given above of $\Bbb C/\overline{\Bbb Q}$ is not an algebraic extension.