Q: Prove that if $\displaystyle H$ and $\displaystyle K$ are normal subgroups of a group $\displaystyle G$, then $\displaystyle HK\leq\\G$.

A: Since $\displaystyle H\leq\\G$ and $\displaystyle k\leq\\G$, $\displaystyle H$ and $\displaystyle K$ have the same identity element $\displaystyle e$. Thus, $\displaystyle ee=e\in{HK}$ which shows $\displaystyle HK$ is not empty.

Now, let $\displaystyle x,y\in{HK}$. Then $\displaystyle x=ab, y=cd$ and $\displaystyle y^{-1}=(cd)^{-1}=d^{-1}c^{-1}$. So,

$\displaystyle xy^{-1}=(ab)(cd)^{-1}=(ab)c^{-1}d^{-1}=a(bc^{-1})d^{-1}$.

Mistake here. Read your line above! In any case, what is the conclusion of this calculation? You want to show that $\displaystyle xy^{-1} \in HK$.
If we let $\displaystyle x=y$, then $\displaystyle xy^{-1}=e\in{HK}$. So, HK is a subgroup of H.

Now, to show it is a normal subgroup, do I just compute $\displaystyle xHK$ and show it is equal to $\displaystyle HKx$?

Yes! You can just use the fact that $\displaystyle xHKx^{-1}=(xHx^{-1})(xKx^{-1})$...
I am not sure if I am showing enough in the proof I wrote down. Some help would be great.

Thanks