Proper normal finite index subgroups There is a well known fact says that if $H$ is a finite index subgroup of $G$ then $H$ has a finite index subgroup $N$ which is normal in $G$.
If $H$ is not normal in $G$ then one can find $N$ proper, namely $N \neq H$.
My question is as follows:
If $H$ is normal in $G$, is there always a proper finite index subgroup $N$ in $H$ which is normal in $G$?
 A: No: it can very well happen that a group has only one nontrivial normal subgroup. The symmetric group $\mathfrak S(n)$, for $n \geq 5$, has this property. 
A: No -- take $G$ to be any infinite simple group, and take $H = G$.  For an especially dramatic example, take a Tarski Monster group.
If you want an example where $H$ is actually proper in $G$, let $G_1$ be any infinite simple group, and take $G = G_1 \times G_2$, where $G_2$ is any finite simple group.  For what you need to know about normal subgroups of a product of two nonisomorphic simple groups, see here.
A: Not in general. Take an infinite simple group $H$ which admits an outer automorphism $a$, and let $G$ be the semdirect product $H \langle a \rangle$ (in fact, your question as stated does not exclude the possibility that $H =G $, in which case taking $H = G$
to be any infinite simple group does the trick, but as I indicate, this sort of thing can still happen when $H$ is a proper normal subgroup of $G$). An explicit example
with $H$ not quite simple is to take $H = {\rm SL}(3,\mathbb{C})$, and $a$ to be the automorphism of $H$ given by $M \to (M^{t})^{-1}$ for each invertible matrix of determinant $1$. This is an outer automorphism of $H$. The only proper normal subgroup of $H$ is its center $Z(H)$, which consists of scalar matrices and has infinite index
(and is normal in $G = H\langle a \rangle$).
