Show if $N$ is normal subgroup of $G$ and $H$ is a subgroup of $G$, then $N \cap H$ is normal subgroup of $H$.

It's way easier (algebraically) than you're making it out to be, but it's tricky to get into the right mindset.

Given that $N \lhd G$ with $H$ some subgroup of $G$, we'd like to show that $N \cap H \lhd H$.

In other words, we need to show that conjugating anything in $N \cap H$ by something in $H$ lands us back in $N \cap H$.

So, let $j \in N \cap H$, and let $h \in H$. You just need to show that $h^{-1}jh$ is in $H$ and $N$ also (hence in $N \cap H$). Think about it in these terms before reading on. Seriously.

Why should conjugating $j$ by $h$ land us back in $H$?

Well, because $j \in H$, and $H$ is a subgroup.

Why should conjugating $j$ by $h$ land us back in $N$?

Well, because $j \in N$ and $N \lhd G$, with $h \in G$; conjugating anything in $N$ by anything in $G$ (of which $H$ is a subgroup) lands us back in $N$.

So, we started with $j \in N \cap H$, and we ended up with $h^{-1}jh \in N \cap H$ so we're done. Easy on algebra, but it definitely takes a certain viewpoint. When I first learned algebra, I would have tried to use way more equations and junk, for what it's worth.