If $H \triangleleft K \leqslant G$, which requirements must be placed on $K$ in order to obtain $N \triangleleft G$? Let $H$ be a normal subgroup of $K$, which is a subgroup of $G$.
By just sketching some computations it seems that $H$ is not necessarily a normal subgroup of $G$ even if $K$ is a normal subgroup of $G$. I haven't proved it since I didn't really know how to structure the proof. 
My motivation for this is understanding $A_n \leqslant S_n$. Most algebra books give the proof for the simplicity of $A_n$ if $n \geq 5$ and another proof, or example, for why $A_n$ is the only non-trivial normal subgroup of $S_n$. This suggests that because $A_n$ is the only non-trivial normal subgroup of $S_n$ does not mean that $A_n$ is simple because there could be normal subgroups of $A_n$ which are not normal subgroups of $S_n$. Is this correct?
 A: Indeed, even if we were able to show that $A_n$ is the only non-trivial normal subgroup of $S_n$ for $n\geq 5$ without using the simplicity of $A_n$, this result alone wouldn't imply that $A_n$ is simple. 
Now coming back to your assertion, assume that :
$$N\triangleleft K\text{ and } K\triangleleft G$$
Then $N$ needs not be normal in $G$. Example :
$$K=S_3^2$$
$$\text{ define a semi-direct product }G:=S_3^2\rtimes_{\phi}\mathbb{Z}/2$$
where for $(\sigma_1,\sigma_2)\in S_3^2$  $\phi(0)(\sigma_1,\sigma_2)=(\sigma_1,\sigma_2)$ and $\phi_1(\sigma_1,\sigma_2):=(\sigma_2,\sigma_1)$. By definition $K\triangleleft G$. 
Clearly $N:=A_3\times \{1\}$ is a normal subgroup of $S_3\times S_3=S_3^2=K$ but you can show that $N=A_3\times \{1\}$ is not normal in $G$ :
$$((1,1),1)\cdot((\sigma_1,1),0)=((1,\sigma_1),0)\notin N $$
What you need is something stronger :
Let $K$ be a group, we say that a subgroup $N$ is characteristic in $K$ if for all $\phi\in Aut(K)$ we have $\phi(N)=N$. 
The proposition you can try to prove is the following if $K\triangleleft G$ and $N$ is a characterisitic subgroup of $G$ then $K$ is normal in $G$.
Hint :

 Let $g\in G$, show that the function $i_{g,N}:N\rightarrow N$ sending $n$ to $gng^{-1}$ is a well defined group automorphism of $N$. Conclude.

A: In general, let $G$ be a non-abelian group, $N$ a simple normal subgroup, and assume that $|G:N|$ is prime and $Z(G)=1$. Then the only non-trivial normal proper subgroup of $G$ is $N$. 
In your case the simplicity of $A_n$ ($n \geq 5$) implies that $S_n$ has only $3$ normal subgroups, $\{(1)\}, A_n, S_n$.
So how do we prove the statement mentioned? First, observe that $G'=N$, since $G/N \cong C_p$, for some prime $p$, and is abelian, we get $G'\subseteq N$. Since $N$ is simple and $G$ is non-abelian, $G'=N$. Now if $M \unlhd G$, then either $M \cap N=1$ or $M \cap N=N$. In the latter case $N \subseteq M$ and using $|G:N|$ being prime we conclude $M=N$ or $M=G$. 
If $M \cap N=1$, this means $M \cap G'=1$ and using that $M$ is normal, this implies $M \subseteq Z(G)$, whence $M=1$.
