# Let $L/K$ be a finite separable field extension. How to prove there are finitely many intermediate fields?

By the Primitive Element Theorem, $L=K(\alpha)$ for some $\alpha \in L$. So, there is a $K$-basis for $L$ given by powers of $\alpha: 1, \alpha,...,\alpha^{n-1}$ where $n$ is the degree of the extension. Let $K \subset F \subset L$ be an intermediate field. Then we may write $F$ as the span of some linear combinations of $\alpha^i$ with coefficients in $K$. By the Steinitz Exchange Lemma, we can choose all but one of the basis vectors to be a power of $\alpha$, and one of them must be 1. Moreover, by the Tower Law $[F:K]$ divides $[L:K]$. How can I proceed from here?

Let $f\in K[X]$ be the minimal polynomial of $\alpha$ over $K$. Let $K\subset K_1 \subset K(\alpha)$ be an intermediate extension. Let $f_1$ be the minimal polynomial of $\alpha$ over $K_1$. Then $f_1\in K_1[X]$ is a factor of $f$. Let's notice that $K_1$ is the field $K_1'$ generated over $K$ by the coefficients of $f_1$. Indeed, the extensions $K(\alpha)/K_1$, $K(\alpha)/K_1'$ have the same degree and $K_1'\subset K_1$.
Therefore, every intermediate extensions of $K(\alpha)$ can generated by an factor of $f$ from $K(\alpha)[X]$. Since there are finitely many such factors, we get finitely many intermediate extensions.
Since $L$ is finite and separable over $K$, the Galois closure $\mathscr L$ of $L$ over $K$ is also finite over $K$. You can construct $\mathscr L$ by taking the minimal polynomial $f$ of your primitive element and adjoin all roots of $f$. In other words, $\mathscr L$ is the splitting field of $f$, and so both separable and normal.
Now appeal to Galois Theory: the fields intermediate between $\mathscr L$ and $K$ are in one-to-one correspondence with the subgroups of the Galois group $\mathbf G^{\mathscr L}_K$. But this is a finite group, has only finitely many subgroups, so there are only finitely many fields between $K$ and $\mathscr L$. Consequently only finitely many between $K$ and $L$.