We're trying to prove that for a tower of fields $K/E/F$, if $a$ is algebraic over $F$ then $[E(a):F(a)] \leq [E:F]$. This is what I have so far,
$\textbf{Proof:}$ Suppose for a contradiction that $[E(\alpha) : F(\alpha)] > [E : F]$. First, we claim that $\alpha$ is algebraic over $F$, it must be algebraic over $E$. Indeed, any polynomial $f \in F[x]$ such that $f(\alpha) = 0$ is contained in $E[x]$. Next, we claim that $[F(a) : E] \leq [E(\alpha) : E]$. Indeed, let $m(x)$ be the minimal polynomial of $\alpha$ in $E[x]$ and $f$ be the minimal polynomial of $\alpha$ in $F[x]$. Then $f(x) \in E[x]$ and hence $m(x) \mid f(x)$. Hence $\deg f(x) \geq \deg m(x)$ so that $[F(\alpha) : F] \geq [E(\alpha) : E]$. We proceed with a case analysis. If $[F(\alpha) : F] = [E(\alpha) : E]$ then by our hypothesis and the tower theorem we would have that $$ [E(\alpha) : F] = [E(\alpha) : F(\alpha)] [F(\alpha) : F] > [E(\alpha) : E][E: F] = [E(\alpha) : F] ,$$ a contradiction. If $[F(\alpha) : F] > [E(\alpha) : E]$, then again by our hypothesis and the tower theorem we would have that $$ [E(\alpha) : F] = [E(\alpha) : F(\alpha)] [F(\alpha) : F] > [E(\alpha) : E][E: F] = [E(\alpha) : F] ,$$ a contradiction. We conclude that $[E(\alpha) : F(\alpha)] \leq [E : F]$
I was wondering if someone can check the proof for correctness?
Yes, I think it is correct! I cannot find any mistakes.