 Solution

For each $g\in G$, conjugation sends a subgroup $H$ to $gH{g}^{-1}$, a subgroup of the same order and hence still proper. Identity conjugation fixes every subgroup, andThese verify the group action laws.

The stabilizer subgroup of $B$ for this conjugation action is its normalizer $N_G(B)$. Since $B\le N_G(B)$, the orbit-stabilizer theorem gives the number of distinct conjugates asAll these conjugate subgroups contain the identity, so counting that element only once givesThe last inequality uses the properness of $B$, so $[G:B]>1$. There is therefore an element outside this union, which is not conjugate to any element of $B$. This proves that conjugates of a proper subgroup do not cover a finite group, including the case $B=\{1\}$.