Suppose $A \subseteq \mathbb R$ is countable. Show $\exists x\in\Bbb R$ s. t.$A \cap (x+A) =\emptyset$. Suppose $A \subseteq \mathbb R$ is countable. Show $\exists x\in\Bbb R$ s. t. $A \cap (x+A) =\emptyset$. (Here, $x + A = \{ x + a : a ∈ A \}$.
I'm unsure how to proceed.    
I thought about taking the smallest distance between the countable numbers of the set etc but I don't think we can take smallest distance of an infinite set even if it's only countable ie say set were rationals.  I've tried by contrapositive, ie suppose there is no such x but that got me nowhere. 
Any hints?

One can show something stronger: let $I$ be uncountable set, then $(f_λ)_{λ∈I}$ a sequence of $I$-many injective functions from uncountable set $X$ to itself such that $f_a(x)=f_b(x)⇒a=b$, then for every countable set $A$, there exists some $i∈I$ such $A\cap f_i[A]=\emptyset$.
Let $I=X=\Bbb R$ and $f_a(x)=a+x$ to get your question.

To prove that let $x∈X$, then if $x\in f_a[A],f_b[A]$ for $a\ne b$, then $f_a^{-1}(x)≠f_b^{-1}(x)$, because $A$ is countable, the set $I_x=\{λ∈I\mid x\in f_λ[A]\}$ is also at most countable, therefore $J=\bigcup_{x∈A}I_x$ is countable union of countable sets, hence countable, so we can take $\kappa\in I\setminus J$, if $A∩f_\kappa[A]≠\emptyset$ there must be $y∈A$ such that $y∈f_\kappa[A]$, so $\kappa\in J$, contradiction.
