Let $f$ be an irreducible separable polynomial of degree $n$ over a field $k$. Let $G$ be the Galois group of the splitting field of $f$ over $k$. Let $G$ be simple. Then, $G$ is a subgroup of $A_n$.
Since $G$ is simple, it has no proper normal subgroup. So no intermediate field between $k$ and the splitting field of $f$ can be normal over $k$.
To show subgroup of $A_n$ I need to show $G$ fixes $\sqrt{D}$ where $D$ is the discriminant. I am not sure how the proof should go.
Suppose $G$ is simple. Then $G$ is isomorphic to a transitive subgroup of $S_n$ (you can see this by seperability, irreducibility and considering the action on the roots). Consider the homomorphism $\phi:G\rightarrow \{1,-1\}^\times$ given by taking the sign of each permutation. This homomorphism has kernel all even permutations in $G$. Hence, all even permutations in $G$ form a normal subgroup. If $G$ is nontrivial, then either:
Case 1: $\phi$ has trivial kernel and $G$ consists only of elements of order $2^k$. From this it is possible to show $G$ is either not simple or is isomorphic to $C_2$ - which is simple and not a subgroup of $A_2$ - the statement is false in this case: look for example at $x^2-2$.
Case 2: $\phi$ has nontrivial kernel. The kernel is a normal subgroup of $G$ and hence must be the whole of $G$.
In either case, $G$ lies in $A_n$ so long as $n>2$.