$f \in K[X]$ of degree $n$ with Galois group $S_n$, why are there no non-trivial intermediate fields of $K \subset K(a)$ with $a$ a root of $f$?
Suggestion for where to start. $M$ stands for the splitting field:
- Identify the subgroup $H=\operatorname{Gal}(M/L)\le\operatorname{Gal}(M/K)\simeq S_n$ as a group of permutations.
- Prove that $H$ is a maximal subgroup of $S_n$, i.e. there are no subgroups properly between $H$ and $S_n$.
- Apply Galois correspondence.
Adding details in response to a request.
Assume that $f(x)\in K[X]$ is a degree $n$ polynomial such that $G=Gal(M/K)\simeq S_n$, $M$ a splitting field over $K$. Let $X=\{\alpha_1,\alpha_2,\ldots,\alpha_n\}$ be the set of zeros of $f(x)$ in $M$. Let $L=K(\alpha_1)$ be an intermediate field. Because any automorphism $\in G$ is determined by the way it acts on $X$, we see that any permutation $\pi\in Sym(X)$ with the property $\pi(\alpha_1)=\alpha_1$ comes from an automorphism of $M$. We can continue to denote it $\pi$, and view it as an element $\pi\in Gal(M/L)$. So $Gal(M/L)$ is thus identified with the point stabilizer $$H=Stab_{Sym(X)}(\alpha_1).$$
Claim. $H$ is a maximal subgroup of $G=Sym(X)$.
Proof. Assume that there were a subgroup $G_1\le G$ properly containing $H$. The set $X$ is partitioned into two orbits of $H$, namely $\{\alpha_1\}$ and the rest of them. If $\sigma\in G_1\setminus H$, then $\sigma(\alpha_1)\neq\alpha_1$. Therefore the two $H$-orbits become a single $G_1$-orbit. In other words, $G_1$ acts transitively on $X$. By the orbit-stabilizer theorem $$ |G_1|=|X|\cdot |Stab_{G_1}(\alpha_1)|. $$ But $H\le Stab_{G_1}(\alpha_1)$ has order $(n-1)!$. Therefore $|G_1|\ge n(n-1)!=n!$. OTOH $G_1\le Sym(X)$, so $|G_1$ can have order $\ge n!$ only when $G_1=Sym(X)$. QED.
Corollary. There are no intermediate fields between $K$ and $L$.
Proof. The Galois correspondence would associate such an intermediate field $E$ with a group $Gal(M/E)$ strictly between $Gal(M/L)$ and $Gal(M/K)$. But we just saw that no such group exists.
The above argument fails in the case $K=\Bbb{Q}$, $f(x)=x^4-3$ because the splitting field $M=K(\root4\of3,i)$ is a degree eight extension of $K$ only. Therefore $Gal(M/K)$ has order eight as well, and thus must be isomorphic to the dihedral group $D_4\le S_4$ of order eight. The point stabilizer of $D_4$ in $S_4$ has order two only (the orbit-stabilizer theorem again). More importantly, that point stabilizer is not a maximal subgroup of $Gal(M/K)$. This allows the existence of intermediate fields between $L=K(\root4\of3)$ and $K$.
But, all this depends heavily on basic results of Galois theory. I don't know of a way of communicating this argument without either referring to Galois theory (or possibly reproducing its relevant parts).
Let $M$ be the splitting field of $f$ and let $a_1,\ldots,a_n$ be the roots of $f$ where $a_1 = a$. Every automorphism of $M|K$ permutes the set $\{a_1,\ldots,a_n\}$ and the fact that the Galois group of $f$ is $S_n$ means that every permutation $\sigma \in S_n$ defines such an automorphism by sending $a_i$ to $a_{\sigma(i)}$. An automorphism $\sigma \in G(M|K)$ fixes $L$ if and only if $\sigma(a) = a$, i.e. $\sigma(1) = 1$ when considered as an element of $S_n$. That means that $G(M|L)$ as a subgroup of $S_n$ is the subgroup $S_{\{2,\ldots,n\}}$ of permutations fixing $1$.
The intermediate extensions $K \subseteq E \subseteq L$ now correspond to the subgroups $G$ of $S_n$ containing $S_{\{2,\ldots,n\}}$, so we have to show that every such subgroup is either $S_{\{2,\ldots,n\}}$ or $S_n$.
If $G$ contains an element $\sigma$ with $\sigma(1) \neq 1$, then it contains all transpositions of the form $$(1\; i) = \sigma^{-1} ( \sigma(1) \; \sigma(i)) \sigma,$$ (unless $\sigma(i) = 1$) and also all transpositions $(i\; j)$ with $i,j \neq 1$. Since $S_n$ is generated by the transpositions it follows that $G = S_n$.
Edit: To show that $(1\; i) \in G$ where $\sigma(i) = 1$ we may assume $n \geq 3$ (the case $n \leq 2$ is trivial). Take any $j$ with $j \neq 1$ and $j \neq i$. Then $(1\;i) = (i\;j)(1\;j)(i\;j) \in G$.