Enumerations of the Rationals and a Refined Denseness Property Let $(y_n)_{n=1}^{\infty}$ be an enumeration of all rational numbers (this is possible, since $\mathbb Q$ is countable) and let $y\in\mathbb R$  be given. Since $\mathbb Q$ is dense in $\mathbb R$, it is clear that for all $\varepsilon>0$, there exists some $n\in\mathbb Z_+$ such that $|y_n-y|<\varepsilon$.
Consider the following strengthening of this property: Given any $y\in\mathbb R$, there exists some $n\in\mathbb Z_+$ such that $|y_n-y|<1/n$, irrespective of the manner in which $\mathbb Q$ is enumerated.
My concern is that it may be possible to enumerate the rationals in such a “fancy” way that the union of a collection of open balls with decreasing radii around them covers only a strict subset of $\mathbb R$, but the statement still feels intuitively true. Even if the statement is false for general enumerations, can one give at least one such enumeration for which this property holds for all $y\in\mathbb R$? What do you think?
 A: Well, the problem amounts to covering $\mathbb{R}$ with family of balls ${B_n}$ with radii $\frac{1}{n}$ and centers at rationals. It's indeed possible to enumerate rationals in "fancy" way - since $[0, 1]$ contains infinitely many rationals, we can "spend arbitrarily much time" there, and, say, for each $2^n$ elements of the sequence place just one outside $[0, 1]$. 
As for the second question: I feel it should be possible, though I suspect it's a little tricky. Will think about it later. 
Edit: Actually, not that tricky after all. 
The idea is based on the fact that we certainly can provide such covering if we drop the requirement to enumerate all the rationals, since $\Sigma_n 1/n$ diverges. Similarily, we can do it with balls half the size. We can thus create the desired enumeration in the following way: let elements with even indices be placed such that their balls cover the $\mathbb{R}$, and let the odd elements enumerate the remaining rationals.
Edit2: It's not quite relevant, but perhaps of interest. I believe more is true: every enumeration of rationals contains subsequence with such property. We can inductively build such subsequence by "growing" the covered area, picking elements from intervals of size $\frac{1}{2n}$ outside the covered region (with some overlap) with index greater than that of any element taken so far. It's possible, because each such interval contains infinitely many elements of enumeration - equivalently, elements with arbitrarily high indices. 
(Such subsequence will not, of course, in general be an enumeration of rationals)
A: For every enumeration $(y_n)$, let $Y_{n}=(y_{n}-1/n,y_{n}+1/n)$ for every $n$.

it may be possible to enumerate the rationals in such a “fancy” way that the union of a collection of open balls with decreasing radii around them covers only a strict subset of $\mathbb R$.

Here is such a way. 
Assume that $|y_n|\leqslant1$ except when $n$ is a power of $2$. Then $Y_n\subseteq[-2,2]$ except when $n$ is a power of $2$ and the length of each $Y_{2^k}$ is $2^{1-k}$ hence $\bigcup\limits_nY_n$ covers at most $[-2,2]$ union a set of measure at most $\sum\limits_k2^{1-k}$, which is finite. In particular, $\bigcup\limits_nY_n\ne\mathbb R$.

can one give at least one such enumeration for which this property holds for all $y\in\mathbb R$?

One can, here is an example.
For every $k\geqslant0$, let $I_k=[-2^k,2^k]$. Let $J_0=I_0$ and, for every $k\geqslant1$, $J_k=I_k\setminus I_{k-1}$. Use $(y_{2n})$ to cover successively $J_k$ for every $k\geqslant0$ by the intervals $Y_{2n}$. Once this is done, use $(y_{2n+1})$ to enumerate the remaining rational numbers.
Then $(y_n)$ enumerates the rational numbers and $\bigcup\limits_nY_{2n}=\mathbb R$.
