Prove or disprove: There is a metric on $R$ with respect to which…?

Here is the text:

For each point $p$ of $\Bbb R$ is $sum_p$ the family of subsets of $\Bbb R$ containing $p$ and can be obtained by deleting from $\Bbb R$ not more than countable infinity of points.

Prove or disprove: There is a metric on $\Bbb R$ with respect to which, for any fixed point $p\in\Bbb R$, every neighborhood of $p$ contains some element of family $sum_p$.

Let $(\Bbb R,d_e)$ be the real field equipped with the Euclidean metric, $r> 0$ a radius and the disk $B(p, r) =\{ x \in\Bbb R : | (p-x)|< r\}$. In the neighborhood of $p$ there are infinitely many points $x$ belonging to a set of $sum_p$. In fact, eliminating a 'multitude of points at most countable, it may remain as a minimum' set $I-A$ which is the set of irrational numbers $I$ minus a countable subset of $A$.

So there is always a neighborhood of $p$ contains some element of family $sum_p$ sum of an infinite set most countable.

-
If I get this right, every open ball $B(p,r)$ should contain a set $A$ such that $\mathbb R\setminus A$ is countable. This is not the case with the standard metric as $B(p,1)$ misses the uncountably many points in $[p+r,p+r+1]$. – Hagen von Eitzen Jan 22 '13 at 15:10

Assume there is such a metric $d$. Let $a=d(0,1)$. Then $B(0,\frac12 a)$ and $B(1,\frac12 a)$ are disjoint by the triangle inequality. By the special property of the metric there is a set $U\in sum_0$ with $U\subseteq B(0,\frac12a)$ and a set $V\in sum_1$ with $V\subseteq B(1,\frac12 a)$. Since $U\cap V$ differs from $\mathbb R$ at most by countably many elements missing from $U$ plus countably many elements missing from $V$, we see that $U\cap V$ differs from the uncountable set $\mathbb R$ by at most countable many elements, hence is not empty. But then also $B(0,\frac12 a)\cap B(1,\frac12 a)\ne\emptyset$, which is absurd.