Sequence of open intervals, decreasing radii going to zero, centered at the rationals might still give all of $\mathbb{R}$? When one first learns measure theory, it is a small novelty to find out that
$$\bigcup_{n=0}^\infty B_{\epsilon/2^n}(r_n)$$
is not all of $\mathbb{R}$, where $\{r_n\}$ is an enumeration of the rationals and $\epsilon$ is an arbitrary positive number (notice this fact is equally impressive if $\epsilon$ is small or large).
Of course, by measure arguments, the set above has measure at most $\epsilon$ and can't be all of $\mathbb{R}$. However, there doesn't seem to be another canonical line of reasoning that explains why the union above is not all of $\mathbb{R}$. That makes me wonder, what if we remove that ability to use this argument?

Is there a pair of sequences of positive real numbers $\{c_n\}$ and $\{d_n\}$ both tending to $0$ such that 
  $$\sum_{n=0}^\infty c_n=\infty=\sum_{n=0}^\infty d_n$$
  where we can demonstrate
  $$\bigcup_{n=0}^\infty B_{c_n}(r_n)=\mathbb{R}\quad\text{and}\quad\bigcup_{n=0}^\infty B_{d_n}(r_n)\neq\mathbb{R}$$
  with a fixed enumeration of the rationals $\{r_n\}$?

An existential proof of both questions would be sufficient for me. But an explicit $\{c_n\}$ and $\{d_n\}$ would be interesting to see.
I feel like the $\{c_n\}$ construction might be fairly easy in comparison to $\{d_n\}$, and using dependent choice, I even think I have an argument off the top of my head: just let $\{c_n\}$ be fairly constant until you swallow up $[-N,N]$ and then let it decrease. Continue ad infinitum. But what about $\{d_n\}$?
 A: Yes, this is possible.  Your proposed construction of $(c_n)$ works with no difficulty.  To construct $(d_n)$, the easiest thing to do is just pick one point that you want to not be covered.  So fix some irrational number $\alpha$.  We would like to just let $d_n=|\alpha-r_n|$.  Then $\alpha$ will not be in any $B_{d_n}(r_n)$, but $\sum d_n$ will obviously be infinite.  This does not satisfy that $d_n\to 0$, but you can easily modify it so that it does (just shrink the $d_n$ so that they converge to $0$ but the sum still diverges).
(This construction of $d_n$ illustrates that it really shouldn't be surprising that an open set can contain all the rationals but not be all of $\mathbb{R}$, since a trivial example of such a set is $\mathbb{R}\setminus\{\alpha\}$!)
A: One approach is to let $c_n=\frac 1n$ and let $d_n=\frac 1n$ when the rational is  outside $(-1,2)$ and $d_n=2^{-n}$ when the rational is inside.  The interval $(0,1)$ then has the measure argument work because none of the balls centered outside $(-1,2)$ can reach there.  We just need to make sure enough of the early rationals are outside $(-1,2)$ to make sure the sum of $d_n$ diverges.  We can just specify that all the rationals in $(-1,2)$ will be in odd positions in the enumeration.  The sum of $\frac 1{2n}$ diverges, so the sum of $d_n$ will diverge.  
To make the union of the $c_n$ balls cover $\Bbb R$ we start with $r_1=-3,r_2=-3-1-0.9\cdot \frac 12,r_3=-1+1+0.9\cdot \frac 13$ with the idea that we make sure to cover a harmonically expanding interval.  Then let $r_4$ pick one of the rationals less than $-3$ in the covered interval and $r_5$ pick up one of the rationals greater than $-3$ in the covered interval.  Continuing, $r_{4k+2}$ is outside the growing interval by $0.9\cdot \frac 1{4k+2}$ on the minus side, $r_{4k+3}$ is outside the growing interval by $0.9\cdot \frac 1{4k+3}$ on the positive side, $r_{4k+4}$ is a fill-in less than $-3$ and $r_{4k+5}$ is a fill-in greater than $-3$.  The union of these balls will covr all of $\Bbb R$
