I wanted to prove:
Let $L/K$ be a simple algebraic field extension. Then $L/K$ has just a finite number of intermediate field.
I worked out a proof and then looked into the solution, which is totally different so I wanted to know if my proof is valid.
The official solution goes like this, let $L = K(a)$ and consider the minimal polynomial $P$ of $a$ over $K$. Let $B$ be an intermediate field, and let $Q$ be the minimal polynomial of $a$ over $B$. Then $Q | P$ in $B[x]$, but $P$ just contains a finite numbers of irreducible factors, and because every intermediate field corresponds to just one factor, it follows that there just could be a finite number of intermediate fields.
My solution: Suppose there exists an infinite number of intermediate fields, then one could construct an infinite tower of field extensions
$$
K_1 \le K_2 \le K_3 \le K_4 \le \ldots
$$
with $[K_{i+1} : K_i] > 1$. And then with the multiplicity and because of $K_i \le L$ for each $i$ it follows that $L/K$ could not be finite. First for $K_1$ set $K_1 := K$, and then proceed inductively. Suppose $K_1 \le \ldots \le K_n$ are choosen, then because there are an infinite number of intermediate fields, there exists some intermediate field $B$ which contains an element $\beta \in B$ which is in none of the of the field $K_i$, i.e. $\beta \notin K_i$ for $i \in \{ 1, \ldots, n \}$. Set $K_{n+1} = K_n(\beta)$, then $K_n \le K_{n+1}$ and $[K_{n+1} : K_n] > 1$ because the inclusion is proper. q.e.d.
EDIT: Added the supposition that $L/K$ has to be algebraic.
Best Answer
This doesn't work. You'd have to prove that $a\notin K_n$ for each $n$.
Of course, I can't provide a counterexample with infinitely many intermediate fields as the theorem is true, but it's not hard to find a counterexample to your reasoning: consider $K={\bf Q}$, $a=\sqrt 2+\sqrt 3$. Then you could pick $K_2={\bf Q}[\sqrt 2]$ and then $B={\bf Q}[\sqrt 3]$ and $\beta=\sqrt 3$ are an intermediate subfield and its element which is not in $K_2$, and then you'd have $K_3=L$.