Let me make my comment into an answer just get things off the ground. I claim that ${\rm PSL}(n,q)$ contains $A_5$ whenever its order is divisible by $60$.
Clearly ${\rm SL}(n,q)$ contains ${\rm SL}(m,q)$ for all $m \le n$, and hence ${\rm PSL}(n,q)$ contains some central quotient of ${\rm SL}(m,q)$ (which is something between ${\rm SL}(m,q)$ and ${\rm PSL}(m,q)$). In particular, ${\rm PSL}(n,q)$ contains a central quotient of ${\rm SL}(5,q)$ for all $n \ge 5$. Since the centre of ${\rm SL}(5,q)$ has order $1$ or $5$, and $A_5$ has no perfect $5$-fold central extension, if ${\rm PSL}(5,q)$ contains $A_5$, then so does ${\rm SL}(5,q)$. Also $|{\rm PSL}(5,q)|$ is divisible by $60$ for all $q$. So it suffices to prove the claim for $n \le 5$.
The result is well-known for $n=2$.
For $n=3$, ${\rm PSL}(3,q)$ contains ${\rm PSL}(2,q)$ as the subgroup ${\rm P}\Omega(3,q)$, and $|{\rm PSL}(3,q)|$ is divisible by $60$ if and only if $|{\rm PSL}(2,q)|$ is, so the result holds.
For $n=4$, ${\rm PSL}(4,q)$ contains the subgroup ${\rm P}\Omega^-(4,q) \cong {\rm PSL}(2,q^2)$ which has order divisible by $60$ for all $q$ and hence contains $A_5$.
For $n=5$, ${\rm PSL}(5,q)$ contains $A_5$ as the image of the natural permutation representation.
Let me carry on with the classical groups. This is not too hard.
Virtually the identical argument works for the unitary groups. In dimensions $3$ and $4$ they contain the same orthogonal groups, and the permutation representation of $A_5$ clearly preserves the identity matrix as unitary form.
For the symplectic case, the obvious containment ${\rm Sp}(m,q) \le {\rm Sp}(n,q)$ for $m \le n$, $m,n$ even projects onto ${\rm PSp}(m,q) \le {\rm PSp}(n,q)$, and ${\rm PSp}(4,q)|$ is divisible by $60$ for all $q$, so it is enough to do it for $n=4$. From the ATLAS, the $4$-dimensional irreducible representation of ${\rm SL}(2,5)$ has indicator $-$ and integer character values, so its image reduces to a subgroup ${\rm SL}(2,5) < {\rm Sp}(4,q)$ for all $q = p^e$ with $p>5$. In fact it works also for $p=5$, and the cases ${\rm PSp}(4,2) \cong S_6$ and ${\rm PSp}(4,3)$ (which contains $2^4:A_5$) can be done separately.
Finally, all of the simple orthogonal groups in dimension at least $5$ contain ${\rm P}\Omega(5,q) \cong {\rm PSp}(4,q)$, so there is nothing new to do there.
Now for some of the smaller rank exceptional groups. The Suzuki groups $^2B(q)$ have order not divisible by $3$. There are containments ${\rm PSL}(2,q) < {\rm SL}(3,q) < G_2(q) < {^3D}_4(q)$, and their orders are all either divisible by $60$ or not, so that covers them. Similarly ${\rm PSL}(2,q) < {^2G}_2(q)$.
That leaves $F_4$, $^2F_4$, $E_6$ $^2E_6$, $E_7$ and $E_8$. For $^2F_4(2^{2n+1})$, I suspect that they all contain the smallest one in the class, the Tits group $^2F_4(2)'$, which (according to the ATLAS) contains $A_5$.
This is probably not the best reference, but I think Table 2 in
http://seis.bristol.ac.uk/~tb13602/docs/large_4.pdf
is enough to deal with the remaining exceptional groups. For example $F_4(q)$ contains $D_4(q)$ = $\Omega^+(8,q)$, which contains ${\rm P}\Omega(5,q)$, which we considered above. There are lots of papers around on subgroups of the exceptional groups, but it's hard to extract exactly what we want.
Best Answer
In an email correspondence with Atiyah, I brought this up. The comments are meant to provide background for Thompson's remark.
"When I first heard of FT I thought there should be a simpler proof using fixed point theorems and K-theory and I propagated the idea. The problem was that fixed point theorems could only deal with fixed points of elements or cyclic groups. So I knew we needed a theory that would cover fixed points of a whole group. We could then apply it to the action of a finite group on the projective space of the reduced regular representation."
He went on further to say, "It was only recently that I realized we had to use equivariant K-theory and not its completion at the identity." Then he mentioned his completion theorem of Brauer induction (with Segal), and how Snaith had given a topological proof of Brauer induction. Although not definitive, Atiyah may have a shortened proof of the Feit-Thompson theorem, and if so it would be presented in the near future.
In case it is helpful, I see that the quote from Gorenstein's paper was around 1960. Atiyah had published the finite group version of the Atiyah-Segal completion in 1961 ("Characters and cohomology of finite groups"). So, based on this recent email correspondence, these 1961 ideas make their appearance in a formulation of FT.