Is the intersection between a subgroup, and a normal subgroup, normal in the parent group? Given $H, N \subset G$ and $N \lhd G$ is there some underlying fact or theorem for why $H \cap N$ would or would not be normal in $G$? My reasoning would be that it would be normal to $G$ as; $\forall x\in H \cap N, x \in N$, and thus it would be normal. Is this sufficent to prove this true or would I need further reasoning?

I know $H \cap N \lhd H$ due to the second isomorphism theorem, could this be extended to G some how?

Thank you for your time.
This is generally false: Consider $G=S_{n}$ for $n\ge3$ and $H=\langle(ab)\rangle=\{e,(ab)\}$
where $(ab)\in S_{n}$ is a two cycle.

$H\leq G$ but is not a normal subgroup (why ?).

Let $A$ be the set of all two cycles in $S_{n}$ and $N=\langle A\rangle$then
$N\trianglelefteq G$ (why ?)

Now 
$$
H\subseteq N\implies H\cap N=H
$$

thus the intersection is not normal in $G$ as $H$ is not.

Note: the fact that $H\cap N\trianglelefteq H$ is simple and we do
not need the second isomorphism theorem - if $h\in H$ then 
$$
h^{-1}Hh=H
$$

since $H$ is a group.

$$
h^{-1}Nh=N
$$

since this is true for all $g\in G$ and in particular to elements
of $h$. thus if $x\in N\cap H$ and $h\in H$ we have it that 
$$
h^{-1}xh\in H\cap N
$$

which proved the normality of $H\cap N$ in $H$.

In general, I don't know about sufficient and necessary conditions
for this to hold. There are, however, some simple sufficient conditions
such as that that the order of $H,N$ are co prime (why ?)
Although your solution is correct, your $N$ is just $S_n$ (although it is true that this is a normal subgroup and the rest of the proof works). To generate $S_n$, we don't even need all two cycles, only $(1 2)$, $(1 3)$, $\dots$, $(1 n)$ is enough (because if you want to move $i$ to $j$, where $i \neq 1$, you only need to apply $(1 i)$ and then $(1 j)$. The remaining element is for $1$).