To say that $p$ is a condensation point of a subset $S$ of a topological space $X$ is to say that any open neighbourhood of $p$ contains uncountably many points of $S$.
Let us suppose we have written $\mathbb{R} = \bigcup_{i=1}^\infty A_i$. I would like to know whether the condensation points of some $A_i$ must form a set with nonempty interior.
My thoughts: First of all, some of the $A_i$ may be countable. But, the union of all such $A_i$ is 1st category in the Baire space $\mathbb{R}$, so the union of all the uncountable $A_i$ is 2nd category in $\mathbb{R}$. Hence there is some uncountable $A_n$ which is not nowhere dense. The difficulty is that we could have $A_n$ equal to, say, the union of the rationals with a Cantor set, in which case the condensation points of $A_n$ are exactly the Cantor set (which has empty interior).
It is my feeling, although I can't seem to prove it, is that there must be at least one $A_i$ out there which makes use of its uncountability to fail to be nowhere dense.
I think the answer to this question is "yes". The key thing I hadn't noticed which makes this easy, pointed by Bryan Scott below, is that if $A \subset \mathbb{R}$ and $C$ is the set of condensation points of $A$, then $A - C$ is countable. As I see it, this is because of the following
Fact: If $S$ is an uncountable subset of a second-countable topological space $X$, then $S$ has a condensation point.
Proof. Otherwise, there is a cover $(U_s)_{s \in S}$ of $S$ such that each $U_s$ contains only countably many points of $S$. Since $X$ is second-countable, $S$ is second-countable, hence Lindelof in the subspace topology. Thus we can extract a countable subcover from $(U_s)$ and this shows, contrary to assumption, that $S$ is countable. QED.
Continuing, if $A-C$ were uncountable, it would have a condensation point which would also be a condensation point of $A$ so $A-C$ must be countable. Since countable implies 1st category, it follows from $\mathbb{R} = \bigcup_{i=1}^n A_i = \bigcup_{i=1}^n (A_i-C_i) \cup C_i$ that for some $i$ the condensation points $C_i$ of $A_i$ do not form a nowhere dense set. Since the condensation points of a subset of a topological space always form a closed set (clearly their complement is open), this $C_i$ is closed and not nowhere dense, or equivalently, is closed with nonempty interior.