About the proof that a simple group of order 60 is isomorphic to A5

I would avoid the counting argument, since there are indeed groups of order $60$ with $15$ Sylow $2$-subgroups, and all elements of order $2$ are shared by several of these Sylow subgroups at once.

Here is a slightly easier way to do this, using a slightly stronger version of Sylow's theorems. Namely,

Theorem: Let $p$ be a prime, and $p^r$ the highest power of $p$ dividing $G$. For any two Sylow $p$-subgroups $P$ and $Q$ of $G$, suppose we have $p^k\le |P\cap Q|$. Then the number of Sylow $p$-subgroups of $G$ - call it $n_p$ - satisfies $n_p\equiv 1\pmod{p^{r-k}}$.

The proof is exactly the same as the usual proof (using group actions), you just have to pay extra attention to orbit sizes along the way.

Now, if $G$ is a group of order $60$, containing $15$ Sylow $2$-subgroups, then since $15\not\equiv 1\pmod{4}$, the theorem implies the existence of two Sylow $2$-subgroups - call them $P$ and $Q$ - with $|P\cap Q|=2$. But then $P\cap Q$ is normal in both $P$ and $Q$, and thus its normalizer, $N_G(P\cap Q)$, has order divisible by $4$, and size at least $|P|+|Q|-|P\cap Q|=6$. Thus $N_G(P\cap Q)$ has size at least $12$.

The action of $G$ on the right cosets of $N_G(P\cap Q)$ gives a homomorphism from $G$ into $S_5$. If we are assuming $G$ is simple, this becomes an isomorphism with $A_5$.

[Note that this is actually a contradiction, since $A_5$ does not have $15$ Sylow $2$-subgroups, but either way, you are done.]