How to show that the normalizer is the largest subgroup of a group in which a set is normal? Assume we have $H \trianglelefteq K\subseteq G$ 
We want to prove the $K \subseteq N_G(H)$
However all I have so far is:
Assume $K \subseteq G$ such that $H$ is normal in $K$. Then it must be that $kHk^{-1} \subseteq H$
If $kHk^{-1} = H$ Then trivially $k \in N_G(H)$, but I do not know what to do if that is not the case.

You have your solution. Since $H$ is normal in $K$, every element of $K$ normalizes $H$ (as in your solution) and hence $K\subseteq N_G(H)$.
