Proof that$ (a+A)\cap A=\varnothing$ if $A$ is countable

Let $B = \{ x \mid (x + A)\cap A \neq \varnothing \} \subseteq \{ x-y \mid x,y \in A \} = C$. Of course $C$ is countable, hence is $B$, but $\mathbb{R}$ in uncountable and therefore $\mathbb{R} \backslash B$ is not empty.

Let $a$ be such that $(a+A) \cap A$ is nonempty, i.e. there are $b,c \in A$ such that $a + b = c$.

Then $a = b-c$ lies in the set $\{b - c : b,c \in A\}$.

Since this set is countable with $A$ being countable, it is not all of $\mathbb{R}$.