(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 $x_n \in K_{i_n}$.  Set $K = \bigcup_n K_{i_n}$.  Since $K$ is meager $K \ne A$, 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.

Added: This indeed shows that a second countable Baire space cannot be the union of a chain of nowhere dense subsets.  Here is a stab at a counterexample in the non-second countable case.

Consider the non-separable complete metric space $\ell^\infty = \ell^\infty(\mathbb{N})$.  I claim its Hamel dimension $\dim \ell^\infty$ is $\mathfrak{c}$.  First, we have the natural inclusion $\ell^1 \subset \ell^\infty$; $\ell^1$ is a separable Banach space, so it is known that $\ell^1$ has Hamel dimension $\mathfrak{c}$, and thus $\dim \ell^\infty \ge \mathfrak{c}$.  On the other hand, $\ell^\infty$ is the continuous dual of $\ell^1$, and thus is naturally included into the algebraic dual of $\ell^1$, which must also have Hamel dimension $\mathfrak{c}$; thus $\dim \ell^\infty \le \mathfrak{c}$.  By Schroeder-Bernstein, $\dim \ell^\infty = \mathfrak{c}$.

Now suppose we believe the continuum hypothesis $\mathfrak{c} = \aleph_1$.  Pick a Hamel basis $B$ for $\ell^\infty$; since it is in bijection with the least uncountable ordinal, we can well-order it in such a way that for any $x \in B$, $B_x = \{y \in B : y < x\}$ is countable.  Note $B$ has no greatest element, so $\bigcup_{x \in B} B_x = B$.  Let $E_x = \mathrm{span } B_x$; clearly $\{E_x\}$ is a chain, and $\bigcup_{x \in B} E_x = \ell^\infty$.  But each $E_x$ has countable Hamel dimension and therefore is separable, so it must be nowhere dense in $\ell^\infty$.