I'm aware that the natural analogue of the Cantor–Schröder–Bernstein (CSB) theorem is not true for fields. That means it is possible for a pair of fields $E$ and $K$ to be such that $E$ is isomorphic to a subfield of $K$, $K$ is isomorphic to a subfield of $E$, but $E$ and $K$ are not isomorphic.
This question is about whether the CSB property holds with respect to a stricter way to compare fields.
Let $E$ and $K$ be fields. Let's say "$E\leq K$" iff $E$ is isomorphic to a subfield of $K$ and $K$ is algebraic over this subfield. More clearly, we say "$E\leq K$" if there is an injective field homomorphism $\phi:E\rightarrow K$ such that $K$ is algebraic over $f(E)$. Henceforth, we will refer to such a field homomorphism as an algebraic homomorphism.
First and main question: Is it true that if $E\leq K$ and $K\leq E$, then $E\cong K$?
Related second question: given a field $E$, if $f:E\rightarrow E$ is an algebraic endomorphism, must we have that $E=f(E)$? That is, must $f$ be an automorphism of $E$?
Counterexample: @Martin Brandenburg points out that for any field $k$, a counterexample to this second question is given by the map $f:k(T)\rightarrow k(T)$ that linearly extends $T\mapsto T^2$. $f$ is a field endomorphism, and $k(T)$ is algebraic over $f(k(T))$.
The answer to the second question is yes in the following special case:
Theorem 1: Let $E$ be a field, and let $f:E\rightarrow E$ be an algebraic field endomorphism; that is to say, $E$ is algebraic over $f(E)$.
Let $F=\{x\in E\:|\:f(x)=x\}$. Then $F$ is a subfield of $E$ that is fixed pointwise by $f$, and $F\subseteq f(E)$. Suppose that $E$ is algebraic over $F$.
In this special case, $E=f(E)$.
Proof:
For each $\alpha\in E$, we want to show that $\alpha\in f(E)$.
Let $\min_F(\alpha)$ be the unique monic irreducible polynomial with coefficients in $F$ of which $\alpha$ is a root, and let $d=\text{deg}_F(\alpha)$ be the degree of $\min_F(\alpha)$.
Let $S=\{f^n(\alpha)\}_{0\leq n\leq d}$, where by $f^n(\alpha)$ I mean $f(f(\cdots f(\alpha)\cdots))$.
Since $F$ is fixed pointwise by $f$, every element of $S$ is a root of $\min_F(\alpha)$. But $\min_F(\alpha)$ only has $d$ roots, whereas $S$ has $d+1$ elements. This means that for some $k>0$, $\alpha=f^k(\alpha)\in f(E)$, as desired. $\blacksquare$
How is the second question related to the first?
Suppose $f:E\rightarrow K$ and $g:K\rightarrow E$ are algebraic field homomorphisms. Then $K$ is algebraic over $f(E)$, so $g(K)$ is algebraic over $gf(E)$. Since $E$ is algebraic over $g(K)$, it follows that $E$ is algebraic over $gf(E)$. If the answer to the second question were "yes", then $E=gf(E)$, and so $g$ and $f$ would be a isomorphisms.
Here is a positive answer to the first question in a special case:
Theorem 2: Let $E$ and $K$ be fields, and let $f:E\rightarrow K$ and $g:K\rightarrow E$ be algebraic field homomorphisms. Let $F\subseteq E$ be defined by $F=\{x\in F \:|\:gf(x)=x\}$. Then $F\subseteq E$ is a subfield. If $E$ is algebraic over $F$, then we must have that $f$ and $g$ are actually isomorphisms between $E$ and $K$.
Proof:
Apply Theorem 1 to see that $gf(E)=E$, and thus $g:K\rightarrow E$ is an isomorphism. Since $gf$ and $g$ are isomorphisms, $f$ must also be an isomorphism. $\blacksquare$
This may seem like a rather useless special case, but this gives us the following nice result:
Corollary: Let $E$ and $K$ be fields. Suppose that $E$ and $K$ are both algebraic over their respective prime fields (their unique minimal subfields). If there are algebraic field homomorphisms $f:E\rightarrow K$ and $g:K\rightarrow E$, then $E$ and $K$ are isomorphic.
You should be able to obtain a counterexample from any pair of isogenous but non-isomorphic elliptic curves: If $C$ and $D$ are such curves over any fixed field $K$, then an isogeny $f:C\to D$ furnishes an algebraic embedding of their function fields $f^*:K(D)\to K(C)$. Moreover, $f$ always comes with a dual isogeny $g:D\to C$, which gives the desired field embedding $g^*$ in the other direction. Hence, $K(D)$ and $K(C)$ are algebraic extensions of each other, which are not isomorphic over $K$ unless $C$ and $D$ are isomorphic curves. By picking $K=\mathbb{Q}$ (or any other prime field), we can thus ensure that the resulting function fields are not isomorphic overall.
Note that these function fields of curves $K(C)$ are of transcendence degree 1 over $K$ and are thus, by your own observation, in some sense minimal counterexamples. Moreover, there can be no counterexample using "simpler" curves, i.e. those of genus zero: Their function fields are all isomorphic to $K(T)$. By Lüroth's theorem, any field $L$ such that $K\subsetneq L\subseteq K(T)$ is of the form $L=K(f(T))$ for some rational function $f$ and therefore isomorphic to $K(T)$ itself.
Edit: Here is an explicit example taken out of Silverman: We take $K=\mathbb{Q}$ and consider the two curves $$C:y^2=x^3+x \hspace{5mm} \text{and} \hspace{5mm} D:Y^2=X^3-4X.$$ These curves are not isomorphic (since -4 is not a sixth power in $K$, see e.g. here) but are isogenous via the map $f:C\to D$ given by $$f(x,y)=\left(\frac{y^2}{x^2},\frac{y(1-x^2)}{x^2} \right).$$ The function fields ($K(C)=K(x,y)$ with $y^2=x^3-x$ and similarly for $K(D)=K(X,Y)$) together with the embedding $$f^*:K(D)\to K(C), X\mapsto \frac{y^2}{x^2},Y\mapsto \frac{y(1-x^2)}{x^2}$$ thus form a concrete instance of the described phenomenon.