Can a countable nowhere dense set have uncountably many accumulation points?

I have a set $S \subseteq \mathbb{R}$ that is countable and nowhere dense. Is it possible that this set has uncountably many accumulation points?

I think the answer is no. I have tried to somehow map the accumulation points to $S$ in such a way that I would only map countably many accumulation points to any point in $S$ but I did not succeed in this.

My question is related to this question. That question drops the assumption of being nowhere dense, after which the answer is yes. However, the example (rationals) are far from being nowhere dense, making me think that the answer to my question is no.

Yes, it can. Call $C$ Cantor's set, which is closed, uncountable and with empty interior. $\Bbb Q\cap C$ is dense in $C$, because it contains all the points of $C$ that have finite ternary expansion. Therefore, $\Bbb Q\cap C$ is an example of what you've asked.

More generally, $\Bbb R$ is a separable metric space, therefore all its subsets have a countable dense subset (i.e. for all $X\subseteq \Bbb R$ there is some countable $S\subseteq X$ such that $\overline S\supseteq X$). Any such subset $S$ coming from a closed uncountable subset $X$ with empty interior will do.