Distinct Sylow $p$-subgroups intersect only at the identity, which somehow follows from Lagrange's Theorem. Why?

That's because it is not true in general. Look at $2$-Sylows in $S_5$: they have nontrivial intersection.

Suppose $P$ and $Q$ are Sylow p-subgroups of prime order p (so not just any power of p; as others remarked, then it is not true in general). Note that $P\cap Q$ is a subgroup of $P$ (and of $Q$). So by Lagrange, the order $|P\cap Q|$ divides p. As p is prime, it is 1 or p. But it cannot be p, as $P$ and $Q$ are distinct. So $|P\cap Q|=1$ and consequently the intersection is trivial.

In some situations, to prove that groups of order $n$ cannot be simple, you can use the counting argument if all Sylow subgroups have trivial intersection, and a different argument otherwise.

For example let $G$ be a simple group of order $n=144 = 16 \times 9$. The number $n_3$ of Sylow 3-subgroups is 1, 4 or 16. If $n_3 = 1$ then there is normal Sylow subgroup and if $n_3= 4$ then $G$ maps nontrivially to $S_4$, so we must have $n_3 = 16$.

If all pairs of Sylow 3-subgroups have trivial intersection, then they contain in total $16 \times 8$ non-identity elements, so the remaining 16 elements must form a unique and hence normal Sylow 2-subgroup of $G$.

Otherwise two Sylow 3-subgroups intersect in a subgroup $T$ of order 3. Then the normalizer $N_G(T)$ of $T$ in $G$ contains both of these Sylow 3-subgroups, so by Sylow's theorem it has at least 4 Sylow 3-subgroups, and hence has order at least 36, so $|G:N_G(T)| \le 4$ and $G$ cannot be simple.