By using some more powerful results, it is possible to do this a lot easier.
The two main ingredients for this will be the following:
Burnside's $pq$-Theorem: If only two distinct primes divide the order of $G$, then $G$ is solvable.
Burnside's Transfer Theorem: If $P$ is a $p$-Sylow subgroup of $G$ and $P\leq Z(N_G(P))$ then $G$ has a normal $p$-complement.
So when looking for a non-abelian simple group, the first result immediately tells us that we can assume at least $3$ distinct primes divide the order of $G$.
The way we will use the second result is the following:
Let $P$ be a $p$-Sylow subgroup where $p$ is the smallest prime divisor of $|G|$. We will show that if $|P| = p$ then $P\leq Z(N_G(P))$ (and then the above result says that $G$ is not simple).
To see this, we note that $N_G(P)/C_G(P)$ is isomorphic to a subgroup of $\rm{Aut}(P)$ (this is known as the N/C-Theorem and is a nice exercise), which has order $p-1$, and since the order of $N_G(P)/C_G(P)$ divides $|G|$, this means that $N_G(P) = C_G(P)$ and hence the claim (since we had picked $p$ to be the smallest prime divisor).
The same argument actually shows that if $p$ is the smallest prime divisor of $|G|$, then either $p^3$ divides $|G|$ or $p = 2$ and $12$ divides $|G|$. The reason for this is that if the $p$-Sylow is cyclic of order $p^2$ the precise same argument carries through, since in that case all prime divisors of $|\rm{Aut}(P)|$ are $p$ or smaller.
If $P$ is not cyclic then the order of $\rm{Aut}(P)$ will be $(p^2-1)(p^2 - p) = (p+1)p(p-1)^2$ and the only case where this can have a prime divisor greater than $p$ is when $p = 2$ in which case that prime divisor is $3$, and we get the claim.
In summary we get that the order of $G$ must be divisible by at least $3$ distinct primes, and either the smallest divides the order $3$ times, or the smallest prime divisor is $2$ (actually, by Feit-Thompson, we know this must be the case, but I preferred not to also invoke that), and $3$ must also divide the order.
This immediately gives us $60$ as a lower bound on the order of $G$, and the next possible order would be $2^2\cdot 3\cdot 7 = 84$ and all further orders are greater than $100$. So we are left with ruling out the order $84$ (which you have done), and showing that the only one of order $60$ is $A_5$.
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 $S_5$.
Best Answer
As was remarked before, you do not have to assume that $G \cong A_5$.
(1) You did that one correctly!
(2) If $|Syl_3(G)|=4$, and $S \in Syl_3(G)$, then $|G:N_G(S)|=4$. Now let $G$ act on the left cosets of $N_G(S)$ by left multiplication, then the kernel of this action is $core_G(N_G(S))=\bigcap_{g \in G}N_G(S)^g$, which is a normal subgroup. Hence, by the simplicity of $G$, it must be trivial and $G$ can embedded in $S_4$, a contradiction, since $60 \nmid 24$.
(3) We prove that if a non-abelian simple $G$, with $|G|=60$, has an abelian subgroup of order $6$, then $G \cong A_5$. This gives a contradiction, since it is easily seen that $A_5$ does not contain any elements of order $6$ (note that an abelian group of order $6$ must be cyclic).
So assume $H \lt G$ is abelian and $|H|=6$. $H$ is not normal so, $N_G(H)$ is a proper subgroup (if not then $H$ would be normal) and since $|G:N_G(H)| \mid 10$, we must have $|G:N_G(H)|=5$ ($=2$ is not possible since subgroups of index $2$ are normal). Similarly as in (2), $G/core_G(N_G(H))$ embeds homomorphically in $S_5$ this time. Of course $core_G(N_G(H))=1$, so $G$ is isomorphic to a subgroup of $S_5$ and since it is simple it must be isomorphic to $A_5$ (if we write also $G$ for the image in $S_5$, consider $G \cap A_5 \lhd G$ and use $|S_5:A_5|=2$).
(4) In general: if $G$ is a group with a unique element $x$ of order $2$, then $x \in Z(G)$. Why? Because for every $g \in G$, $g^{-1}xg$ has also order $2$ and must be equal to $x$. In your case $G$ is non-abelian simple, so $Z(G)=1$.
So only (1) is the true statement.
Edit For case (3) I forgot the case where $|G:N_G(H)|=10$. I have a proof that is quite sophisticated and maybe there is an easier way.
Anyway, in this case $H=N_G(H)$. Consider the subgroup $P$ of order $3$ of $H$. This must be a Sylow $3$-subgroup of $G$, since $3$ is the higest power of $3$ dividing $|G|=60$. Observe that in fact $N_G(P)=H$. This follows from what we showed in (2): $|Syl_3(G)|=|G:N_G(P)|=10$ and of course $H \subseteq N_G(P)$.
Trivially, $P \subset Z(N_G(P))$. Now $P$ satifies the criterion of Burnside's Normal $p$-Complement Theorem, see for example Theorem (5.13) here. But then $P$ has a normal complement $N$, such that $G=PN$ and $P \cap N=1$. Now $G$ is non-abelian simple, so $N=1$ or $N=G$, which both lead to a contradiction.