What's wrong with my proof? nZ are the only subgroups of Z 
Show that if $H$ is a subgroup of $(\mathbb Z,+)$ then $\exists n\in\mathbb N$ such that $H=n\mathbb Z$.

And here's what I did:
Suppose that H is a subgroup of Z and $\forall n\in\mathbb N$, $H \neq n\mathbb Z$.
then
$\forall n\in\mathbb N$ ($\exists x \in H$ and x $\notin n \mathbb Z $) or  ($\exists x \in  n\mathbb Z$ and x $\notin H $)
suppose $\forall n\in\mathbb N$ ($\exists x \in H$ and x $\notin n \mathbb Z $)
for n=|x|we have x $\in H$ and x $\in |x|\mathbb Z$ which is absurd.
suppose $\forall n\in\mathbb N$ ($\exists x \in  n\mathbb Z$ and x $\notin H $)
since x $\in \cap n \mathbb Z$ then x=0 which is $\in$H and this is absurd.
When I showed this proof to my recitation teacher, he said that the proof is wrong, but he couldn't find why.

You wrote "$\forall n\in \Bbb N \, \exists x\in H$" so this $x$ depends on $n$ and it isn't universal which means that for $n,p\in \Bbb N$ there are $x_n, m_p\in H$ where $x_n$ and $x_p$ may be different.
