We let, $$D_{60} = \langle r, s: r^{30} = e, s^2 = e, rs = sr^{-1} \rangle.$$ The subgroups of order $2$ in $D_{60}$ are the cyclic groups generated by $s,r^{15},sr,sr^{2},\dots,sr^{14}$. So there are a total of $14 + 2 = 16$ subgroups of order $2$ in $D_{60}$. For $A_{5}$, we simply have to find subgroups of cycle type $3,2$. There are $\binom{5}{3}\times \binom{3}{2}\times \frac{1}{2} = 15$; hence, these two groups cannot be isomorphic.
Does my proof look good? Any mistakes? Thanks.
It looks nice. Another simplier approach would be that $A_5$ (Abel's lemma) is simple, hence non solvable, and $D_{60}$ as $\langle r^{15}\rangle $ as normal subgroup