Let $K$ be closed subset of $\mathbb{R}$ with empty interior. Prove that there exists $a \in \mathbb{R}$, such that set $\{a+x: x\in K\}$ is included in set of irrational numbers.
Well, I know that Baire theorem is what should solve that problem. My problems stem from that I am confused how $K$ can look like. Surely we cannot have all irrational numbers included in it. But can we have countably many of them? I suspect we cannot, probably because it would contradict with the fact (no ontological obligations here) that it has empty interior. But what are implications, if any, of that?
Your intuition is incorrect. $K$ can contain infinitely many - even uncountably many! - irrationals and still be a closed set with empty interior (and I'm not sure what "ontological obligations" means in this context).
For an example of a closed set with empty interior containing infinitely many irrationals, consider $\{z+\pi: z\in\mathbb{Z}\}$. This can be modified to be compact, in addition, without much effort.
For an example of a closed set with empty interior containing uncountably many irrationals, think of the Cantor set - although note that it also contains lots of rationals. Coming up with a closed set with empty interior containing uncountably many irrationals and not containing any rationals is a harder task, which I leave as an exercise.
For solving the problem, here's a hint: for $q\in\mathbb{Q}$, let $A_q=\{r\in\mathbb{R}: q\not\in K+r\}$. Then:
How many sets of the form $A_q$ are there?
What kind of set, topologically, is $A_q$ (open? closed? neither?)? Use the fact that $K$ is closed.
What kind of set, distribution-wise, is $A_q$ (dense? not dense? nowhere dense?)? Use the fact that $K$ has empty interior.
Finally, what does the Baire category theorem say about $\bigcap A_q$, and how does that solve your problem?