Theorem. There is no chain of nowhere dense subsets of $\mathbb{R}$ whose union contains an interval.
Proof. Suppose there was such a chain $\{\ B_i \mid i\in I\ \}$, where $\langle I,\lt\rangle$ is a linear order and $i\lt j$ implies $B_i\subset B_j$. First, I claim that this chain cannot have countable cofinality, since then we could find a countable cofinal subset of $I$, and the union of the $B_j$ from this cofinal subset would also contain an interval, violating the Baire category theorem. So every countable subset of $I$ is bounded. In this case, consider the set $Q$ of rational numbers $q$ in the interval from the union $\bigcup_i B_i$. Each of them appears in some $B_{i_q}$, and the set of all $i_q$ for $q\in Q$ is a countable set and hence bounded in $I$. Thus, there is some $j\in I$ beyond all $i_q$. So $B_j$ contains all those $q$ and thus is not nowhere dense. QED