It's the follow metric: $d(x,y)= ||x|| +||y||$ if $x$ and $y$ don't lie on a line through the origin. And otherwise $d(x,y)= ||x-y||$.
I think the answer is no, because I tried it with $\mathbb{Q}^{2}$ as countable and that didn't work. But I don't know how to prove it that it isn't true in general.
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HINT: You're right: it's not. You can prove it by finding an uncountable family of pairwise disjoint, non-empty open sets. I've added a further hint in the spoiler-protected block below.
Consider open rays leaving the origin.
Brian M. Scott
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Yes , I'm sorry, I reversed the conditions. – bob Jun 10 '15 at 10:36
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Ok, thanks. But I don't understand the hint that good. Could you explain it a bit more? – bob Jun 10 '15 at 10:54
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@jens: Did you also look at the hidden hint? – Brian M. Scott Jun 10 '15 at 11:01
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yes I did. But I have to prove that there is no countable dense set, why I have to find an uncountable family? – bob Jun 10 '15 at 11:39
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@jens: Suppose that $D$ is a dense set. Every one of those open sets has to contain a member of $D$. Can $D$ then be countable? – Brian M. Scott Jun 10 '15 at 11:47
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No, it can't. But all these rays, are they uncountable? How do you know? – bob Jun 10 '15 at 12:02
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@jens: There's a ray for each real number in the interval $[0,2\pi)$. (Think polar coordinates.) All non-trivial intervals of real numbers are uncountable. – Brian M. Scott Jun 10 '15 at 12:08
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Ah yes of course , ok. And then its proofed? – bob Jun 10 '15 at 12:14
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@jens: Yes, at that point you have all of the necessary ingredients. – Brian M. Scott Jun 10 '15 at 12:18
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Can you also consider just points on these rays and because D is dense, on every ray there have to be a point of D near that point. And because there are uncountable isolated points, D must uncountable. But aren't they isolated? – bob Jun 10 '15 at 12:32
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@jens: There are no isolated points in this space. – Brian M. Scott Jun 10 '15 at 12:34