Given a countable partition of the reals, must the condensation points of some part have nonempty interior? 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.
 A: For each $i \in \mathbb{Z}^+$ let $C_i$ be the set of condensation points of $A_i$; then $A_i \setminus C_i$ is countable. Let $C_0 = \bigcup\limits_{i \in \mathbb{Z}^+} (A_i \setminus C_i)$; $C_0$ is countable, and $\bigcup\limits_{i\in\omega}C_i = \mathbb{R}$. Clearly some $C_i$ with $i>0$ must be somewhere dense in $\mathbb{R}$. However, it’s entirely possible for all of the $A_i$ themselves to have empty interior, even if each consists entirely of condensation points.
Enumerate $2^\omega \times \omega \times \omega$ as $\{\langle \alpha_\xi,m_\xi,n_\xi \rangle:\xi \in 2^\omega\}$, let $\mathbb{R} = \{x_\xi:\xi < 2^\omega\}$, and let $\{I_n:n \in \omega\}$ be an enumeration of the open intervals with rational endpoints. Recursively construct sets $A_n$ for $n\in\omega$ as follows. Suppose that at stage $\xi < 2^\omega$ the approximations $A_n(\eta)$, each of cardinality $|\eta|$, have been constructed for each $n\in\omega$ and $\eta<\xi$. For $k\in\omega\setminus\{m_\xi\}$ let $A_k(\xi)=\bigcup\limits_{\eta<\xi}A_k(\eta)$. Let $\zeta$ be minimal such that $x_\zeta \in I_{n_\xi}\setminus \left(\bigcup\limits_{k\in\omega\setminus\{m_\xi\}}A_k(\xi)\cup\bigcup\limits_{\eta<\xi}A_{m_\xi}(\eta)\right)$, and let $A_{m_\xi} = \{x_\zeta\}\cup\bigcup\limits_{\eta<\xi}A_{m_\xi}(\eta)$. This is always possible, since $|I_{n_\xi}|=2^\omega>\left|\bigcup\limits_{k\in\omega\setminus\{m_\xi\}}A_k(\xi)\cup\bigcup\limits_{\eta<\xi}A_{m_\xi}(\eta)\right|$. The construction goes through to $2^\omega$, and for each $n\in\omega$ we let $A_n = \bigcup\limits_{\xi<2^\omega}A_n(\xi)$. The sets $A_n$ partition $\mathbb{R}$, and since each of them intersects every non-empty open interval with rational endpoints (in a set of power $2^\omega$, in fact), all of them have empty interior.
