 Solution

The identity commutes with every $h\in H$, so the centralizer $C_G(H)$ is nonempty. If $x,y\in C_G(H)$, then for every $h\in H$,Thus it is closed under multiplication. From $xh=hx$, multiplication by $x^{-1}$ on both sides gives $h{x}^{-1}=x^{-1}h$, so $x^{-1}\in C_G(H)$. These are exactly the subgroup conditions; henceThe proof works for the centralizer of a subset as well; it does not require that $H$ itself be a subgroup.