Is the composition series $1 \leq A_3 \leq S_3$ in $S_3$?
I thought this was because the composition factors are isomorphic to $\mathbb{Z}_2$ and $\mathbb{Z}_3$ and are therefore simple. Also the subgroups are normal. $A_3$ is normal in $S_3$ since it has index $2$?
$1 \le A_3 \le S_3$ is indeed a composition series of $S_3$.
A subnormal series is a composition series if all of its composition factors are simple, and as you noted, the composition factors are isomorphic to $Z_2$ and $Z_3$, and therefore simple groups.
The trivial group is obviously normal in $A_3$, and $A_3$ is normal in $S_3$ because it has a index of 2, as you noted. therefore this is indeed a proper subnormal series.
This proves that this is a composition series.