If {c(n)} is an arbitrary sequence of irrational numbers converging to 0 then Q + c(n), the set obtained by adding c(n) to the set of rational numbers Q, is clearly disjoint from Q for each n. 


Is there an uncountable dense set of the real numbers, say D, for which a sequence {c(n)} converging to 0 exists such that D + c(n) does not intersect D for all n?


If such a set exists, and is borel, then it must have measure 0 since D - D would contain an interval if D had a positive measure.



disjoint translates of a dense uncountable set
Choose your favorite c(n), and define $D$ as any complementary $\mathbb{Q}$-vector space of $\mathrm{span}(\{c(n): n\in\mathbb{N}\})$ in $\mathbb{R}$ as $\mathbb{Q}$ vector space. This $D$ is not measurable, of course (countably many translates of it cover $\mathbb{R}$, so it can't be of measure zero, and on the other hand as you said it can't be of positive measure). It's a version of the construction of the Vitali set (actually the standard Vitali set already works with a sequence of rationals c(n) ).


