If $K\subseteq E$ is an algebraic field extension and $\overline{E}$ is the algebraic closure of $E$, then the field extension $K\subseteq\overline{E}$ is also algebraic.
This is simply stated in my notes but I don't know how to prove it. I have seen this answer by egreg but it just says "by well-known results", which I don't get.
If $a\in\overline{E}$ then it is a root of some polynomial with coefficients in $E$, and every element of $E$ is a root of a polynomial with coefficients in $K$, but I don't know how to use this last statement.
Fact. Let $L/K$ be a field extension, and let $\alpha \in L$. $\alpha$ is algebraic over $K$ if and only if $K(\alpha)/K$ is a finite extension.
You should make sure you can prove this! Let me know if you're stuck.
Now let $\alpha \in \overline{E}$ be arbitrary. Then there is some polynomial $p \in E[x]$ such that $p(\alpha) = 0$. Let $p = \sum_{i=0}^n a_i x^i$. Letting $F = K(a_0, \dots, a_n)$, we see that $p \in F[x]$. Now (exercise!) $F/K$ is finite, and we have that $\alpha$ is algebraic over $F$ so $F(\alpha)/F$ is finite. Finally, $$[K(\alpha) : K] \leq [F(\alpha) : K] = [F(\alpha) : F] [F : K]$$ shows that $K(\alpha)/K$ is finite. Thus, $\alpha$ is algebraic over $K$.