Cover $(0, +\infty )$ by open sets 
Cover $(0, +\infty)$ by open sets $U_\alpha$ such that for any $\epsilon > 0$ there are points $x, y \in (0, +\infty)$ with $|x-y|<\epsilon$, not both belonging to the same $U_\alpha$

The distance beteen $x$ and $y$ is $\epsilon$ so would it work if we took our open setss $U_\alpha=B_{\frac{\epsilon}{2}}(\alpha)=\{x \in \mathbb{R}| d(x, \alpha) <\frac{\epsilon}{2} \}$
So $|x-y|<\epsilon$ but we can take our open ball as having a radius$\frac{\epsilon}{2} $
I feel like I may have missed something - is my answer correct?

Your definition of $U_{\alpha}$ is not clear, is $\alpha$ a point or is it an index? 
Besides that, you want to be able to find these two points for every $\epsilon$.
If I understand correctly, what you need is a family of sets such that you can always find two points very close to each other such that they correspond to different $U_{\alpha}$'s. You can go overkill and try $U_{x,\epsilon}:=B_{\epsilon}(x)\cap (0,+\infty)=\{y\in (0,+\infty)\mid d(x,y)<\epsilon\}$, for every point $x\in (0,+\infty)$ and every $\epsilon >0$.
So now let $\epsilon >0 $, take any $x\in (0,+\infty)$ and the neighborhood $U_{x,\frac{\epsilon}{4}}$ and $y:=x+\frac{\epsilon}{2}$ and the neighborhood $U_{y,\frac{\epsilon}{4}}$, you know that $d(x,y)=\frac{\epsilon}{2} < \epsilon $ and that $U_{x,\frac{\epsilon}{2}}\cap U_{y,\frac{\epsilon}{4}}=\emptyset$
