This is Exercise 32.34 in Fraleigh's book, 'A first course in abstract algebra'.
Show that if $E$ is an algebraic extension of a field $F$ and contains all zeros of every $f(x) \in F[x]$, then $E$ is algebraically closed field.
Below is my proof which I want to get verified. I'm quite certain with my proof, but since this is the first time for me to study algebra, it feels like I'm missing something.
Suppose that there exist polynomial $g(x) \in E[x]$ which has no zero in $E$. Then by Kronecker's Theorem, there exists field $K$ such that $E \le K$ and $K$ contains zero of $g$, which I will denote by $\alpha$. Then we have $F \le E \le E(\alpha) \le K$, since $E(\alpha)$ is the smallest field which contains both $E$ and $\alpha$. Since $E$ is an algebraic extension of $F$ (by assumption) and $E(\alpha)$ is an algebraic extension of $E$ (because $g(\alpha)=0$), $E(\alpha)$ is an algebraic extension of $F$. Thus, $\alpha$ is a zero of polynomial in $F[x]$. By assumption in question, we obtain $\alpha \in E$, a contradiction.
Thank you for reading my question. Any comments (including non-mathematical advices) will help me really lot.