Is this proof of the second isomorphism theorem correct? I remade a proof of the second (some books call it third) isomorphism theorem that is not the same I was taught and I'd like you to check the details as it is much shorter than the one I had.

Let $G$ be a group and $N \trianglelefteq G$, $N \le H \le G$ then:
  
  
*
  
*$H/N \trianglelefteq G/N \iff H \trianglelefteq G $
  
*In that case $\frac{G/N}{H/N} \cong G/H$
  

My proof is the following:
Let $x \in H$ and $y \in G$ then consider the following equality:
$$(yN)(xN)(yN)^{-1} = (yN)(xN)(y^{-1}N) = (yxy^{-1})N$$
we can read this in two directions if we assume $H/N \trianglelefteq G/N$ the equality read from the left gives us that $H \trianglelefteq G $. Reading it the other way gives the other implication.
For the second point let define $f:G/N \rightarrow G/H$ such that $xN \mapsto xH$. I say this is well defined as if $x_1N = x_2N \implies x_1x_2^{-1} \in N \le H \implies x_1x_2^{-1} \in H \implies x_1H = x_2H$. 
Also, $\text{Ker}(f) = \{xN \in G/N:xH=H\} = \{xN \in G/N:x \in H\} = H/N$. So I conclude by the first isomorphism theorem that $\frac{G/N}{H/N} \cong G/H$.
 A: Ad 1) There has to be $x \in H$ and $y \in G$, otherwise the argumentation would not make sense. To prove "$\Leftarrow$" you should also mention that $H/N$ is a subgroup of $G/N$. (That is required in the characterisation of normal subgroups.) To show that, you just have to explain that $H/N = \pi(H)$, hence it is a subgroup of $G/N$, where $\pi \colon G \to G/N$ denotes the canonical projection. [Since $\pi$ is surjective, we even get immediately that it's a normal subgroup, that would even shorten your proof.]
Ad 2) You should verify that $f$ is a homomorphism of groups - otherwise the proof would fail. Then, since your group G is not said to be commutative, you have to argue slightly different: $$x_1 N = x_2 N \Rightarrow x_2^{-1}x_1 \in N \Rightarrow x_2^{-1}x_1 \in H \Rightarrow x_1 H = x_2 H.$$
Again, you here have the chance to shorten your proof: The canonical projection $\pi \colon G \to G/H$ is a homomorphism of groups. By the universal property of quotients (which you use to proof the first isomorphism theorem) there exists a (unique) homomorphism of groups $\phi \colon G/N \to G/H$ which maps $xN$ to $\pi(x) = xH$. That's what you called $f$.
