My attempt at a proof.
Suppose $A_n$, for n $\geq 5$, has a proper subgroup $H$ and $[A_n:H] = m < n$. Kumbuka: $A_n$ is simple for all n $\geq$ 5 a fact that we gonna use later. Take left cosets on H to be $\{aH | a \in A_n\} = C$ a map $\phi : C \rightarrow S_2$ defined by $\phi(aH) = (12)$ if $a \neq (1)$ and$\phi((1)H) = (1)$ .
Checking that $\phi$ is a homomorphism. Give $a, b, \in A_n$ we got $\phi(aHbH) = \phi(abH) = ab = \phi(aH)\phi(bH)$ and by definition $\phi((1)H) = (1)$.
Anyway since $A_n$ is simple $Ker(\phi) = A_n$ so $\phi(aH) = (1) \text{ } \forall \text{ } a \in A_n$ but $\phi((1)H) = (1)$ so we got $aH, (1)H, \in Ker(\phi)$ For sure then $aH = H \text{ } \forall \text{ } a\in A_n$ which implies that $A_n \subset H$ but also $H \subset A_n$ meaning $A_n = H$. This is a contradiction since we supposed that $H$ is a PROPER subset of $A_n$. This means our supposition is dead wrong and $A_n$ got no proper subset of index $m < n$ for $n \geq 5$.
Any feedback is welcome. I'm still sorta new to proof writing for algebra.