Let $K \leq E$, $\mathcal{A}_{E|K}=\{a \in E \text{ with } a \text{ algebraic } |K\}$
$K \subseteq \mathcal{A}_{E|K} \subseteq E$
We claim that $\mathcal{A}_{E|K}$ is a field.
$a, b \in \mathcal{A}_{E|K}$ are algebraic $|K$, so $K \leq K(a,b) \subseteq E$, where $K \leq K(a,b)$ is an algebraic extension.
So, $K(a,b) \subseteq \mathcal{A}_{E|K}$
$K(a,b)$ is a field $\Rightarrow a-b, ab, a^{-1} \in K(a,b) \subseteq \mathcal{A}_{E|K}$
So, we have that $K \leq \mathcal{A}_{E|K} \leq E$ and the extension $K \leq \mathcal{A}_{E|K}$ is algebraic from the definition of $\mathcal{A}_{E|K}$.
Could you explain to me how we conclude that $K(a,b) \subseteq \mathcal{A}_{E|K}$ ??
$K(a,b) = (K(a))(b) \Rightarrow K(a)(b)$ is an algebraic extension of $K(a).$ Also $K(a)$ is an algebraic extension of $K.$ So $K(a, b)$ is an algebraic extension of $K.$ In particular, every element of $K(a, b)$ is algebraic over $K.$ (everything is happening inside $E.$)