The problem asks to find two different algebraically closed fields $\mathcal{E}$ and $\mathcal{F}$ with $\mathcal{E} \subseteq \mathcal{F}$.
We have not done a whole lot of stuff with algebraically closed and in fact the only one that came to mind was $\mathbb{C}$.
Does $\mathbb{C}(x)$, the field of rational functions in $x$ with coefficients from $\mathbb{C}$ work? Our definition of an algebraically closed field (which I'm not sure if there are other standard definitions or not) is that every polynomial with coefficients from the field must have a root in the field. Since the set of polynomials in $\mathbb{C}(x)$ is just $\mathbb{C}[x]$ it seems like my example is algebraically closed as well. Does this in fact work or am I missing something?
Thanks
Edit: I did notice a flaw in my thinking. I need to think of polynomials with coefficients in $\mathbb{C}(x)$ not polynomials contained in $\mathbb{C}(x)$.
The algebraic closure of $\Bbb Q, \overline{\Bbb Q} \subset \Bbb C$ gives an example to your main question.
More generally, for every (edit : u uncountable) cardinality, there is a unique algebraically closed field (upto isomorphism) So while $\Bbb C \subset \overline{C(X)}$(the algebraic closure of $C(X)$) are another pair of examples, the two fields are isomorphic although they are not equal.
(The uniqueness requires the axiom of choice.)