In the above proof, I think when the author wrote $[F(u):F][F:K],$ he already assumed that $F(u)$ is a field. Since $F(u)$ is isomorphic to $F[x]/<f(x)>,$ we know that $F[x]/<f(x)>$ is then a field. However, that implies that $f(x)$ is irreducible over $F,$ which is exactly what we need to prove! Is that a circular argument? I am wondering why we can assume $F(u)$ is a field?
A priori it may not be the case that $F(u)\cong F[X]/\langle f(X)\rangle$. What is certainly the case is that $F(u)\cong F[X]/\langle g(X)\rangle$ where $g$ is the minimum polynomial of $u$ over $F$. Then $g(X)$ is certainly a factor of $f(X)$. The aim is to prove that $g(X)=f(X)$. Anyway, I'm sceptical of what the book says about various degrees are less than various other degrees. I would complete the argument as follows: the degree $|F(u):K|$ is divisible by both $|F:K|$ and $|K(u):K|=\deg f$. As these numbers are coprime, $|F(u):K|=|F(u):F||F:K|$ is divisible by $|F:K||K(u):K|$ and so $|F(u):F|$ is divisible by $|K(u):K|=\deg f$. But $|F(u):F|=\deg g\le \deg f$.