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?
 A: $$\mathscr U:=\{B_{1/n}(n):n\in\mathbb N\}\cup \{(n,n+1):n\in\omega\}$$ is an open cover of $(0,\infty)$. 
Now let $\epsilon>0$. There exists $n\in\omega$ such that $1/n<\epsilon/3$. Then $n-\epsilon/3$ and $n+\epsilon/3$ are distance less than $\epsilon$ apart but are not in any single member of $\mathscr U$.
I approached the problem by thinking about simple open covers - open covers consisting of intervals. I noted that such open cover cannot have a finite subcover if we want your property. So then I thought about my favorite such open cover of $\mathbb R$, which is all the sets $(n,n+1)$ and $(n−1/2,n+1/2)$ (pictured in my head). Then I just thought about shrinking the $(n−1/2,n+1/2)$  intervals so that I can always get close-together $x$ and $y$ at the "ends" of one of them.
A: Your cover can't be defined with a fixed $\epsilon$, because every pair $x$, $y$ with $|x-y|<\epsilon/4$ can be found inside one of your $U_\alpha$.
One solution: A (countable) cover is the collection $U_1, U_2, \ldots$ defined by $U_n:=(n, n+1+\frac1n)$.
A: Cover $C_1=\{A_1,A_2\}$ where $A_1=(0,2)$ and $A_2=(1,\infty).$ 
Cover $C_2=\{A_n: n\in N\}$ where $A_1=(0,2)$, and $A_n=(n-1,\infty)$ for $n\geq 2.$
Cover $C_3=\{A_n :n\in N\}$ where $A_n=(n-1,n+1).$
For all 3 covers :For any $\epsilon >0$  let $e=\min (1,\epsilon).$  Then $x=1-e/3$ belongs only to $A_1$ and $y=1+e/3$ belongs only to $A_2.$ And $|x-y|=2 e/3<\epsilon.$ 
$C_1$ is a finite cover, $C_2$ is an infinite cover, $C_3$ is an infinite cover with no finite sub-cover.
A: It is possible to choose intervals $\{I_1, I_2, \ldots\}$ such that each $I_n$ has length $1/n$ and $\bigcup I_i = \mathbb R$. This relies on how the harmonic series diverges, or in other words
$\displaystyle \frac{1}{2} + \frac{1}{3} +  \frac{1}{4}+ \ldots = \infty$.
So we should be able to cover the real line with that infinite length. 
A: 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 $U_{x,\frac{\epsilon}{2}}\cap U_{y,\frac{\epsilon}{4}}=\emptyset$
