(Joel's answer appeared as I was typing this.)

I think the answer is no.

Suppose to the contrary there exists a nonmeager set $A \subset \mathbb{R}$ which is the union of some chain $\{K_i\}_{i \in I}$ of nowhere dense sets.  $A$ is separable, so we may enumerate a countable dense set $\{x_n\} \subset A$.  Then we can find an increasing sequence $\{K_{i_n}\}$ with $x_n \in K_{i_n}$.  Setting $K = \bigcup_n K_{i_n}$, by the Baire category theorem $K \ne A$, so there exists $x \in A \backslash K$.  Now there must be some $K_j$ with $x \in K_j$.  Now for each $n$ we certainly don't have $K_j \subset K_{i_n}$, so we must have $K_{i_n} \subset K_j$ since the $K_i$ are a chain.  Thus $K \subset K_j$, but then $K_j$ contains all the $x_n$ and so is not nowhere dense.