= Solution
For each $g\in G$, conjugation sends a subgroup $H$ to $gHg^{-1}$, a subgroup of the same order and hence still proper. Identity conjugation fixes every subgroup, and
$$
g_1(g_2Hg_2^{-1})g_1^{-1}=(g_1g_2)H(g_1g_2)^{-1}.
$$
These verify the <group action> laws.
The <stabilizer subgroup> of $B$ for this <conjugation action> is its <normalizer> $N_G(B)$. Since $B\leq N_G(B)$, the <orbit-stabilizer theorem> gives the number of distinct conjugates as
$$
\boxed{m=[G:N_G(B)]\leq[G:B].}
$$
All these conjugate subgroups contain the identity, so counting that element only once gives
$$
\left|\bigcup_{g\in G}gBg^{-1}\right|\leq1+m(|B|-1)\leq1+[G:B](|B|-1)=|G|-[G:B]+1<|G|.
$$
The last inequality uses the properness of $B$, so $[G:B]>1$. There is therefore an element outside this union, which is \b[not conjugate to any element of $B$]. This proves that <conjugates of a proper subgroup do not cover a finite group>, including the case $B=\{1\}$.
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