 Conjugates of a proper subgroup do not cover a finite group

For a proper subgroup $B$ of a finite group $G$, the number $m$ of its conjugates is $[G:N_G(B)]\le [G:B]$, because its normalizer contains $B$. All conjugates contain the identity, so their union has size at most $1+m(|B|-1)\le |G|-[G:B]+1<|G|$. Thus some element is not conjugate to any element of $B$. Finiteness is essential to this cardinal counting argument.