Let $G$ be a finite group.
$(1)$ If $G$ is nilpotent, then $\forall k \in \mathbb{Z} : k \mid |G|$, $\exists H \lhd G : |H| = k$.
$(2)$ If $\forall k \in \mathbb{Z} : k \mid |G|$, $\exists! H \leq G : |H| = k$, then $G$ is cyclic.
For $(1)$, I know that if $G$ is nilpotent, then all Sylow-p subgroups are normal in $G$ and $G$ is isomorphic to the direct product of the Sylow-p subgroups. I know that $P_{p_1}, P_{p_2} \lhd G \implies P_{p_1}P_{p_2} \lhd G$. But what if $P_{p_1} \in{\rm Syl}_{p_1}(G) : |P_{p_1}| = p_1^\alpha : \alpha > 1$, how do I construct a normal subgroup of order $p_1$?
For $(2)$, I am stuck.
As you noted, you can reduce to the case where $G$ is a finite $p$-group. Every finite $p$-group is nilpotent, so it is indeed equivalent to only think about finite $p$-groups.
For (1), we need to show that every finite $p$-group has a normal subgroup of every possible order; i.e. if $\lvert G \rvert = p^k$, then $G$ has a normal subgroup of order $p^\ell$ for all $0 \leq \ell \leq k$. Assuming $k > 0$, recall that $Z(G)$ is nontrivial, and every subgroup of $Z(G)$ is normal in $G$. This gives us a normal subgroup of order $p^\ell$ for all $0 \leq \ell \leq \log_p \lvert Z(G) \rvert$. Now work by induction and think about $G/Z(G)$...
For (2), it turns out that (for any group $G$ whatsoever), whenever $G/Z(G)$ is cyclic, then $G$ is abelian. You should prove this, and work again by induction to show that $G$ must be a direct product of cyclic groups. How can you then conclude that $G$ is cyclic?