Well I do have a small understanding issue here.
If you say that for every $\epsilon > 0$ there are two values $a, b \in \mathbb{R}$ with $a \neq b$ such that $ a - b < \epsilon$. Thats not a problem there.
But when you say that there are two fixed values $x,y \in \mathbb{R}$ such that $x-y < \epsilon$ for every $\epsilon > 0$. How does this not imply that $x = y$ ?
I mean in the upper case you can choose $a = b+\frac{\epsilon}{2}$ or something. But in the second case you already got two fixed values which are so close together that the difference is smaller than every $\epsilon > 0$. We had that in class and the lecturer told as it does not necessary mean that $x = y$.
If, for instance, $x=0$ and $y=1$, then it is true that, for every $\varepsilon>0$, $x-y<\varepsilon$. However, $x\neq y$.
You would be right if the condition was $|x-y|<\varepsilon$.