Is this proof about normal subgroups and quotient groups ok? I proved the following:
Let $N$ be a normal subgroup of $G$ and let $H$ be a subgroup of $G$. If $N$ is a subgroup of $H$ prove that $H/N$ is a normal subgroup of $G/N$ if and only if $H$ is a normal subgroup of $G$.

Please could someone check my proof and tell me if it is correct?

Proof:
$\implies$: Let $H$ be normal in $G$. Then $gH = Hg$ for all $g \in G$. Consider an arbitrary $hN \in H/N$ and an arbitrary $gN \in G/N$. Then
$$ gN hN g^{-1}N = ghg^{-1}NNN = ghg^{-1}N = h' N \in H/N$$
where the last equality follows because $H$ is normal. 
$\Longleftarrow$: Now assume $H/N$ is normal in $G/N$ so that $gN hN g^{-1}N = h' N$. 
Let $h \in H$ and $g\in G$ and let $e$ be the identity. Then
$$ ghg^{-1} = gh nn^{-1}g^{-1}= gn h (gn)^{-1} =  gn he (gn)^{-1} = h'n'$$
for some $n' \in N$ and $h' \in H$. The last equality follows because $he \in H/N$ and $H/N$ is normal in $G/N$.
 A: You don't need to write "N" more than once in each equation-- if you have an expression like hNgN, you should just write it as hgN, since we already know that's how multiplication works in a quotient group and it's much more readable. In the second part of your proof, you seem to have assumed that n commutes with h, so ghn = gnh, which is not true in general. 
I'm on my phone, so let k be the inverse of g. For any n in N, you can simply write ghkn = h'n' for some h' in h and n' in n, by assumption of normality of H/N. Move n to the other side of the equation and note that since N is a subgroup of H, the element on the right side is an element of H. 
A: I wouldn't use the $gH=Hg$ property, once it has been employed for proving that $G/H$ can be given a group structure under $(xH)(yH)=(xy)H$.
What we can use is that $\pi\colon G\to G/N$, $\pi(x)=xN$, is a surjective homomorphism.
Suppose $H$ is normal in $G$. Let $g\in G$ and $h\in H$; then
$$
\pi(g)\pi(h)\pi(g)^{-1}=\pi(ghg^{-1})\in H/N
$$
Suppose $H/N$ is normal in $G$. Let $g\in G$ and $h\in H$; then
$$
\pi(ghg^{-1})=\pi(g)\pi(h)\pi(g)^{-1}\in H/N
$$
so
$$
ghg^{-1}\in\pi^{-1}(H/N)=H
$$

The first part of your proof is good. The second part is wrong, because you're not saying what's $n$.
Suppose $H/N$ is normal in $G/N$. Then, from
$$
(ghg^{-1})N=(gN)(hN)(g^{-1}N)\in H/N
$$
we know that $(ghg^{-1})N=h'N$, for some $h'\in H$. Therefore
$$
ghg^{-1}=h'n
$$
for some $n\in N$. Since $N\subseteq H$, we have $h'n\in H$.
