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 $A_5$.
What you need to show is that for $H$ and $K$ different Sylow 2 subgroups, $H\cap K=<1>$.
Once this is established, there couldn't be 15 Sylow 2 subgroups -- we'd get too many elements together with your determination of numbers of Sylow 3 and Sylow 5 subgroups. So there are 5 Sylow 2 subgroups. Since H is not cyclic, there are exactly 3 elements of order 2 in H, for a total of 15 in G.
Back to $H\cap K=<1>$. Suppose $H\cap K\neq <1>$. Then $H\cap K=<x>$ where $x$ is of order 2. Then $C_G(x)$ has order greater than 4. Use Sylow to argue that this is impossible.
Holt has already answered why a Sylow 2 subgroup can not be cyclic. Here's an expansion of his answer:
In any permutation group if the cycle $c=(i_1,i_2,\cdots ,i_n)$ has length n, the parity of $c$ is $(-1)^{n-1}$. So $c$ is an odd permutation (not a member of the alternating group) if and only if $n$ is even. Furthermore, the parity of a product is the product of the parities.
Suppose G is a group and a Sylow 2 subgroup of G is cyclic of order 4, say the order of G is 4m with m odd. Then G is not simple:
Consider G as a subgroup of the symmetric group on the elements of G (Cayley representation). Let $g\in G$ be of order 4 and $x\in G$. Then the cycle of $g$ containing $x$ is $(x,xg,xg^2,xg^3)$. Thus the cyclic decomposition of $g$ consists of m 4 cycles. By the above paragraph, the parity of $g$ is $((-1)^3)^m=-1$. So $g\notin A_{4m}$. So $S_{4m}=GA_{4m}$ ($A_{4m}$ has index 2). Hence $$S_{4m}/A_{4m}\cong{A_{4m}\over A_{4m}\cap G}$$. Then $A_{4m}\cap G$ is a normal subgroup of G of index 2.
Best Answer
A Sylow $2$-subgroup contains fifteen non-identity elements. Two distinct Sylow $2$-subgroups could conceivably meet in an order $8$ subgroup, so it is difficult to count how many elements are contained in the union of the Sylow $2$-subgroups.
But there are either $1$ or $3$ Sylow $2$-subgroups. In the latter case $G$ acts on them transitively by conjugation, so there is a non-trivial homomorphism from $G$ to $S_3$.