Consider the set of all $2^n$-th roots of unity, as $n$ ranges over the non-negative integers. An infinite subgroup involves elements of arbitrarily high order, which generate everything below them.
What you are looking for is the idea of control of fusion. Let $S$ be a Sylow $p$-subgroup of $G$, and let $H$ be a subgroup of $G$ containing $S$. We say that $H$ controls fusion in $G$ with respect to $S$ if, whenever $A$ and $B$ are two subsets of $S$ that are conjugate in $G$ via some element $g$, then there exists $h\in N_G(S)$ such that $A^h=B$ and, furthermore, conjugation by $g$ and $h$ induce the same function on $A$.
Burnside proved that if $S$ is abelian then $N_G(S)$ controls fusion in $S$ with respect to $G$. In general, if $N_G(S)$ controls fusion then $N_H(T)/C_H(T)$ is isomorphic to a subgroup of $N_G(S)/C_G(S)$, whenever $H$ is a subgroup and $T$ is a Sylow $p$-subgroup of $H$. So your second claim holds whenever $N_G(S)$ controls fusion. This is really a statement about $p$-groups: a $p$-group $S$ where $N_G(S)$ always controls fusion when $S$ is a Sylow in $G$ are called resistant. Almost all finite $p$-groups are resistant.
Counterexamples to your general claim, that $\mathrm{icn}(H)\leq \mathrm{icn}(G)$, abound for simple groups, particularly for the prime $2$, when $\mathrm{icn}(G)$ can equal $1$. In this case, Frobenius's normal $p$-complement theorem states that $G$ has a normal $p$-complement if and only if $\mathrm{icn}(H)$ is a $p$-group for all subgroups $H$. If we choose a group without a normal $2$-complement, but with $N_G(S)=SC_G(S)$, then we should be able to find a counterexample.
For a concrete one, let $G=\mathrm{SL}_2(9)$ and $H=\mathrm{SL}_2(3)$ for $p=2$. Then $\mathrm{icn}(G)$ has order $8$, because $N_G(S)=S$ and the group is quaternion. The Sylow $2$-subgroup of $H$ is also quaternion but of order $8$. However, it is normal in $H$, so $\mathrm{icn}(H)=|H|/2=12$.
Best Answer
Or you could argue as follows. Suppose that $N \not \subseteq Z(G)$. Let $M$ be a maximal normal subgroup of $G$ containing $N.$ Then $M = C_{G}(N)$ as $M$ is Abelian and $M$ is proper, but $N \not \subseteq Z(G).$ Then certainly $M = C_{G}(M).$ Hence $M$ is in fact a maximal subgroup of $G,$ for if $M$ is contained in another proper subgroup $H$ of $G$ normal or not) then $H$ s Abelian, so $H \subseteq C_{G}(M) = M$ and $H = M.$ Hence $G/M$ is a simple group (as $M$ is maximal normal) with no proper non-identity subgroup (as $M$ is not contained in any proper subgroup of $G$). But $G/M$ contains a cyclic subgroup of prime order (you don't even need Cauchy's theorem to see this), so $G/M$ is in fact cyclic of prime order.