Theorem: The subgroups of $\mathbb{Z}$ are $n \mathbb{Z}$ for $n=0,1,2....$.
My Proof: Let $H$ be an arbitrary subgroup of $\mathbb{Z}$. Let $x \in H$. If $x<0$ then since $H$ is closed under taking additive inverses, it follows that we can find a positive element in $H$, hence the subset of $H$ with positive integers is non-empty. Let $X$ be the smallest positive integer in $H$. Now, it suffices to show that $H \subset \langle X\rangle$. So let $h \in H$, by the division algorithm, $h=Xq+r$ where $0\leq r< X$. This shows that $h-Xq=r$. Since $h-Xq=h+(-Xq)=r$, it follows that $-Xq,h$ are members of $H$, and since $H$ is closed, $r$ is in $H$. However, since $r<X$ it follows that $r=0$ hence $h=Xq$ and so every element $H$ is a multiple of $X$ and so $H=\langle X\rangle$ is the set of integers that are multiples of $X$. The case for $x>0$ follows similarly and the case for $x=0$ is trivial..
Is the proof correct? How do I improve it?