**Problem: The space $X$ = (({$\frac{1}{n} | n \in$ $\mathbb{Z}$ \ {$0$}) $\cup$ {$0$}) $\subset \mathbb{R}$ and $\mathbb{Z} \subset \mathbb{R}$ are not homeomorphic in their standard topologies (but $X$ is the continuous one-to-one image of $\mathbb{Z}$).**

I am not sure whether I am correct or wrong in this, but I did it in this way:

$X$ doesn't contain {$0$}, and its not open in $X$. Also, $\mathbb{Z}$ is discrete, so let $h: X \to Z$ is a continuous one-to-one and onto mapping. Let, $h(0) = x$, since {$x$} is open, therefore, its inverse $h^{-1}${$x$} = {$0$} is also open in $X$. Hence, there doesn't exist any continuous one-to-one and onto map between $X$ and $\mathbb{Z}$. 

Please help me give a better proof with proper explanation and arguments for this. Also, notation and style should be best. Appreciate your help and support.