Claim: Suppose $H\le G$ and $P$ is a Sylow $p$-subgroup of $G$. Show that, without reference to Sylow's theorems, there exists some conjugate of $P$ whose intersection with $H$ is a Sylow $p$-subgroup of $H$.
If this is proved, then we can prove Sylow's first theorem by induction backwards: Each group is embedded in a symmetry group, hence embedded in $GL(n,\mathbb{F}_p)$ for some $n$. The latter has a Sylow $p$-subgroup, namely, the subgroup $U$ of all upper unitriangular matrices.
I want to prove the claim by group actions, but I cannot figure out how.
We study the action of $H$ on $X = G/P$. Since $(|X|,p) = 1$, it follows that the number of classes in some orbit $\mathcal{O}$ in $X$ under this $H$-action is relatively prime to $p$. If $H_x \subseteq H$ denotes the stabiliser of any element $x \in \mathcal{O}$, then, check that $H_x$ has the desired form, so that $H_x$ is a $p$-group and $([H: H_x], p) = 1$ so that $H_x$ is a Sylow $p$-subgroup of $H$.