[Math] Proof that a finite separable extension has only finite many intermediate fields

Question

Let $E/F$ be finite separable extension. Is there any proof of the fact that there are only finitely many intermediate fields without using primitive element theorem or fundamental theorem of Galois theory?

Answer 1

I think of the following argumentation:
1) Let $K$ be a field and $n\in \mathbb{N}$. Then $K^n$ has only finite many commutative unital subalgebras. The exact number is $B(n)$ - the $n$th Bell number.
2) Let $(K;L)$ be a separable extension. Then $L$ is separable as $K$-algebra. Thus, by tensoring a suitable splitting field $T$ for $L$ we get that $L\otimes T$ is isomorphic to $T^n$ for a suitable $n$. Hence - using 1) - $L\otimes T$ has only finite many commutative unital subalgebras.
3) The function $A\mapsto A\otimes L$ is injective defined on the commutative unital subalgebras of $L$.
4) Any commutative unital subalgebra of $L$ is a field, too.
Remark: Such algebras are called futile (with only finite many subalgebras) in the literature.