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 $K_3=L$.
I see a couple of problems with your proof. But we will manage to salvage the idea of your proof:
"Problem": The statement that $M\subset M'\Leftrightarrow f_M\in M'[X]$ needs a proof (unless you know this from somewhere). The implication "$\Rightarrow$" is immediate, the other direction is not. The problem is, that it could happen that $M\not\subset M'$ and $M'\not\subset M$.
Solution: If $f_M\in M'[X]$, then $f_M = f_{M\cap M'}$. Let $F:=M\cap M'$ Then we can conclude by a degree argument:
$$[L:K] =[L:F]\cdot[F:K]= [L:M]\cdot[M:F]\cdot[F:K]$$
But $[L:M]=[L:F]$ (since the minimal polynomials are the same) and hence $[M:F]=1$. Thus $M=F=M\cap M'$ and $M\subset M'$.
Problem: From your assumption that $f_M\in M'[X]$ you can't deduce that $f_M=f_{M'}$. As a counterexample look at any nontrivial extension $K(\alpha)/K$. $f_K\neq f_{K(\alpha)}$ since $f_{K(\alpha)}=X-\alpha$. The problem in your argument is the sentence: "However, since both polynomials are irreducible and monic...". Both are only irreducible as polynomials over their respective base fields.
Solution: Strike this paragraph. As you mention in the first paragraph: "Now I want to show that any $f_M$ uniquely determines the corresponding intermediate field $M$." Do just that! Assume that $f_M=f_{M'}$ and you can skip a couple of sentences in your proof.
Problem: Your claim $f_K = \prod_{K \subseteq M \subseteq L}f_M$ doesn't hold. I assume that again you are making the mistake of assuming that all of those $f_M$ are irreducible over $L[X]$? You can check that this is wrong with any non-trivial example. If $f_K = f_K\cdot g$, then $g=1$, so all the $f_M$ in your product would have to be units, thus constant).
Solution: You don't need this. It will suffice that in a UFD any element only has (up to associates) only finitely many divisors (why?). Any $f_M$ divides $f_K$, so there can only be finitely many such $f_M$. Thus there can only be finitely many intermediate fields.
Bonus: Why does any $f_M$ divide $f_K$? We know that $f_K\in M[X]$ and by definition (or a basic lemma, depending on your source) $f_M$ divides any polynomial $g\in M[X]$ that satisfies $g(\alpha)=0$. We know that $f_K(\alpha)=0$.
Best Answer
Based on @JyrkiLahtonen's comment above.
If either $[K_1 : K] = \infty$ or $[K_2 : K] = \infty$, then the inequality is trivially true. So, let both, $K_1/K$ as well as $K_2/K$ be finite extensions. Let $\{ a_1,\dots,a_n \}$ be a basis for $K_1$ over $K$ and let $\{ b_1,\dots,b_m \}$ be a basis for $K_2$ over $K$. Consider $V = \operatorname{span}_K\{ a_i b_j : 1 \leq i \leq n, 1 \leq j \leq m \}$, the vector subspace of $L$ spanned by the vectors $a_i b_j$ over the field $K$. We wish to show that $V$ is a field.
To show that $V$ is a ring, it suffices to show that $a_i b_j \cdot a_k b_l \in V$ for all $1 \leq i, k \leq n$, $1 \leq j, l \leq m$. Now, $$ a_i b_j \cdot a_k b_l = a_i a_k \cdot b_j b_l $$ and $a_i a_k \in K_1$, $b_j b_l \in K_2$. So, we can express them as a $K$-linear combinations of $ a_1,\dots,a_n$ and $b_1,\dots,b_m$, respectively. That is, $$ a_i a_k = \sum_{p = 1}^n c_{ikp} a_p \quad \text{and} \quad b_j b_l = \sum_{q = 1}^m d_{jlq} b_q $$ for some scalars $c_{ikp}, d_{jlq} \in K$. Hence, $$ a_i b_j \cdot a_k b_l = \left( \sum_{p = 1}^n c_{ikp} a_p \right) \cdot \left( \sum_{q = 1}^m d_{jlq} b_q \right) = \sum_{p=1}^n \sum_{q=1}^m (c_{ikp}d_{jlq}) a_p b_q \in V. $$ Hence, $V$ is a ring. In particular, $V$ is an integral domain because it is contained in $L$ which is a field.
Next, we need to show that the multiplicative inverse in $L$ of every non-zero element in $V$ lies in $V$ itself. Let $r \in V$, $r \neq 0$. Since $V$ is spanned over $K$ by a finite set, $V$ is a finite-dimensional vector space over $F$. If $\dim_K V = d$, then the set $\{ 1, r, r^2, \dots, r^d \}$ is a $K$-linearly dependent set. Hence, there exist $c_0,c_1,\dots,c_d \in K$, not all zero, such that $$ c_0 + c_1 r + c_2 r^2 + \dots + c_d r^d = 0. $$ Let $k = \min\{ 0 \leq i \leq d : c_i \neq 0 \}$. Then, $$ c_k r^k + c_{k+1} r^{k+1} + \dots + c_d r^d = 0.\tag{1} $$ It cannot be that $c_i = 0$ for all $i \neq k$ because otherwise we would have $$ c_k r^k = 0 \implies r^k = 0 \implies r = 0, $$ which is a contradiction. Note that here we are crucially using the fact that $V$ is an integral domain. So, we have concluded that $k < d$. Now, from $(1)$ we get that $$ \begin{align} & &c_k r^k + c_{k+1}r^{k+1} + \dots + c_d r^d &= 0 \\ &\implies &r^k(c_k + c_{k+1}r + \dots + c_d r^{d-k}) &= 0 \\ &\implies &c_k + c_{k+1}r + \dots + c_d r^{d-k} &= 0\\ &\implies &r(c_{k+1} + c_{k+2} r + \dots + c_d r^{d-k-1}) &= -c_k\\ &\implies &-c_k^{-1}(c_{k+1} + c_{k+2} r + \dots + c_d r^{d-k-1}) &= r^{-1}. \end{align} $$ So, $r^{-1}$ lies in the $K$-span of $\{ 1 , r, r^2, \dots, r^d \}$ which is a subspace of $V$. Therefore, $r^{-1} \in V$ for all nonzero $r \in V$. Thus, $V$ is a field.
Any field containing both $K_1$ and $K_2$ must contain $a_i b_j$ for all $1 \leq i \leq n$, $1 \leq j \leq m$. Hence, it must also contain the $K$-span of $\{ a_i b_j \}$. But we have just shown that this is a field, so it must be the minimal field containing both $K_1$ and $K_2$. In other words, $L = K(K_1,K_2) = V$. Thus, any basis for $L$ over $K$ can contain no more than $nm$ elements. In other words, $$ [L : K] \leq [K_1 : K] [K_2 : K]. $$ Hence, proved.