I have searched the site for some information on this problem, but the results aren't illuminating enough.
Let $G$ be a simple group of order $60$. Since it is simple, then there must be $6$ Sylow $5-$subgroups in $G$. Conjugation by elements of $G$ give rise to a permutation representation $\varphi:G\rightarrow S_6$ since all Sylow $5-$subgroups are conjugate. Since $G$ is simple, then the $\operatorname{ker}\varphi=\{1\}$. Then by the first isomorphism theorem for groups $\varphi(G)\cong G\le S_6$.
Since $G$ is simple, then $\varphi(G)$ is simple, so that $G\cong\varphi(G)\le A_6$, the largest simple subgroup of $S_6$.
Is this sufficient? Or is there a better way to argue this?
I have hard time understanding why $G\le A_6$ just because $A_6$ is the largest simple subgroup of $S_6$.
I'd argue differently, by looking at the signature homomorphism: $$G\to S_6\to C_2.$$ That must be trivial, otherwise $G$ has a subgroup of index $2$. Thus $G$ is a subgroup of $A_6$.