I don't think so.

Let be the set in question. We may assume WLOG that (since it won't affect the cardinality). (where is the open ball of radius around ). Then, I claim that . To see this let then since we see and by the Archimedean principle there exists some such that . Thus, and since it's trivial that the conclusion follows. But, each is the intersection with of the complement of a bounded neighborhood of and thus countable. So, is the countable union of countable sets and thus countable. But, this contradicts the assumption that is uncountable.