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.
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) ).
Use of the Axiom of Choice is unnecessary here. One can write down an explicit $\mathbb{Q}$-linearly independent subset $\mathcal{T}$ of $\mathbb{R}$ with size $2^{\aleph_0}$ as I wrote in this answer. If it so happens that some of the $c(n)$ are in the $\mathbb{Q}$-span of this set $\mathcal{T}$, then remove the finitely or countably many elements which are involved in expressing the $c(n)$ as a $\mathbb{Q}$-linear combination of elements of the original set $\mathcal{T}$; what remains of $\mathcal{T}$ still has size $2^{\aleph_0}$. The $\mathbb{Q}$-span of this remainder has the desired property. The set in question will be a Borel set (of Lebesgue measure 0).
Yes. We proceed using the axiom of choice. Choose an basis of $\mathbb{R}$ with cardinality $|\mathbb{R}/\mathbb{Q}|$ and a bijection between this basis and the elements of $\mathbb{R}/\mathbb{Q}$. Pick an element from each coset of $\mathbb{Q}$ contained in the corresponding basis set (we may do this as each coset is dense); this will be our set $D$. It is dense as each basis set has non-empty intersection with $D$. Now any sequence of rationals tending to zero works as $c(n)$, as if $D+c(n)$ had non-empty intersection with $D$, then two elements of $D$ would differ by a rational.
