I think the $S_5$ work is correct (sorry, haven't looked at the $A_5$ work). I would do it a somewhat different way:
Since 5, but not 25, divides 120 (the size of $S_5$), the Sylow-5 subgroups of $S_5$ must be cyclic of order 5. There are 24 5-cycles in $S_5$, 4 of them in each of these subgroups, so, 6 Sylow-5 subgroups.
Similarly, the Sylow-3 subgroups must be cyclic of order 3. There are 20 3-cycles in $S_5$, 2 to a subgroup, so 10 Sylow-3 subgroups.
Since 8, but not 16, divides 120, the Sylow-2 subgroups must have order 8. Now, $S_4$ contains three copies of the dihedral group of order 8, and $S_5$ contains 5 copies of $S_4$, so I get 15 Sylow-2 subgroups.
Something like this ought to work for $A_5$.
EDIT: I think OP wants me to elaborate on the dihedral-group part of the argument.
Take a square, label its vertices, cyclically, with 1, 2, 3, 4. Then the element $(1234)$ of $S_4$ has a natural interpretation as the rotation, one-fourth of the way around, of the square, and the element $(13)$ is the flip in the diagonal through 2 and 4, so these two elements of $S_4$ generate a subgroup isomorphic to the dihedral group of order 8.
The same is true for the elements $(1342)$ and $(14)$, and also for the elements $(1423)$ and $(12)$, and those are the three copies of the dihedral group in $S_4$.
Now if you pick any one of the numbers 1, 2, 3, 4, 5, and consider all the elements of $S_5$ that fix that number, you get a subgroup of $S_5$ isomorphic to $S_4$. Those are your five copies of $S_4$ in $S_5$.
It seems to me that there is a significant overlap with the standard proofs, e.g., see here. "Proof that there is only one simple group of order sixty, isomorphic to the alternating group of degree five": The key idea is to prove that $G$ has a subgroup of index five. After that, we use the fact that $A_5$ is simple to complete the proof. Again, Burnside is used. For another idea to use a stronger version of Sylow see here:
About the proof that a simple group of order 60 is isomorphic to A5
So to your question Has anyone seen this proof before? Yes, up to minor variations, which are always possible. In general, this theorem has been treated so often, that almost everything has been said. I am not sure that it is interesting to come up with yet another variation. On the other hand, it is certainly useful to write this up for oneself, and for further topics in group theory.
Best Answer
If $n_2=5$ let $P_1,.., P_5$ be the 5 2-Sylow groups.
Recall that $G$ acts on the 2-Sylov subgroups via conjugation: $P \to xPx^{-1}$.
Now, each $x \in G$ defines a permutation of $P_1,P_2,.., P_5$.
Show that this is an embedding of $G$ into $S_5$.