Any countable $A \subseteq \mathbb{R}$ satisfies $(x+A) \cap A = \emptyset$ for some $x$

Note that $x+A$ and $A$ meet if and only if there are $y,z\in A$ with $x+y=z$, so $x=z-y$. This means that $(x+A)\cap A\ne\emptyset$ iff $x\in A-A$, which is a countable set, as it is the image of $A\times A$ under the map $(y,z)\mapsto y-z$.

So, since $\mathbb R$ is uncountable, we can find values of $x$ not in $A-A$, and any such $x$ works.

Suppose the conclusion does not hold: i.e. for every $x$ in $\mathbb{R}$, there are elements $a_x$ and $b_x$ of $A$ such that $x+a_x=b_x$. This defines an injection from $\mathbb{R}$ to $A^2$: just map $x$ to $(a_x,b_x)$. Thus $|\mathbb{R}|\le |A^2|$, so $|A|\ge |\mathbb{R}|$, and in particular $A$ is uncountable.