$S_8$ inside $A_n$

I think this works. As you say, it is easy to prove this result using the simplicity of $A_{n+1}$ for $n \ge 4$ .

$S_n$ contains an element (the transposition $(12)$ , for example) that has order $2$ and whose centraliser has order $2 (n-2)!$ . If $S_n$ could be embedded injectively in $A_{n+1}$ , then $A_{n+1}$ would contain an element of order $2$ with a centraliser with at least $2(n-2)!$ elements.

If $a$ is an element of $A_{n+1}$ of order $2$ , then $a$ is a product of an even number of disjoint transpositions $t_1,t_2,\ldots,t_k$ , (where $k$ is even). Since $\pi(i\;j)\pi^{-1} = (\pi i \; \pi j)$ for any permutation $\pi$ and distinct values $i,j$ , any element of the centraliser of $a$ in $A_{n+1}$ must act as a permutation on the transpositions $t_1,t_2,\ldots,t_k$ by conjugation, and hence the centraliser of $a$ in $A_{n+1}$ must contain exactly $k! (n-2k)!\times 2^{k-1}$ elements. To see this, there are $k!$ possible permutations of the $k$ transpositions. Once we know, for example, that $\pi (a\;b)\pi^{-1} \,=\, (c\;d)$ , then either $\pi a = c$ , $\pi b = d$ or else $\pi a = d$ , $\pi b = c$ . That gives us another $2^k$ choices. But half of the resulting permutations will be odd.

Since $k$ is even and $2k \le n+1$ , it is easy to see that $k! (n-2k)!\times 2^{k-1} \,<\, 2(n-2)!$ , at least for $n \ge 5$ . Thus, for $n \ge 5$ , $A_{n+1}$ contains no element of order $2$ which has a centraliser with $2(n-2)!$ elements or more. The cases of smaller $n$ can be handled by inspection.