I previously asked a version of this question on Math.SE, but didn't receive an answer. (But there is a bounty there if you want to claim it!)
Let $X$ be a Banach space. (If it helps, feel free to assume that $X$ is separable.) In this question, "subspace" means a linear subspace of $X$, not necessarily closed.
Q1. Suppose $E \subset X$ is a subspace which is meager, so that we can write $E = \bigcup_n E_n$, where the $E_n$ are nowhere dense subsets of $X$. Can the $E_n$ be taken to be subspaces of $X$? That is, can a meager subspace always be written as a countable union of nowhere dense subspaces?
Note that the trivial approach of replacing each $E_n$ with its linear span does not work, since the linear span of a nowhere dense set is not necessarily nowhere dense (consider the unit sphere).
Q1a. If the answer to Q1 is yes, can the subspaces $E_n$ be taken to be increasing, i.e $E_1 \subset E_2 \subset \cdots$? Can we write $E$ as a a countable increasing union of nowhere dense subspaces?
The trivial approach of replacing $E_n$ by $E_1 + \dots + E_n$ does not work, since the sum of two nowhere dense subspaces may not be nowhere dense (for instance, in $C([0,1])$, consider the mean-zero functions and the constants).
This came up in the context of thinking about the following related question.
Q2. Let us say a subspace $E \subset X$ determines weak-* convergence if for every sequence $\{f_n\} \subset X^*$ such that $f_n(x) \to 0$ for every $x \in E$, we have $f_n(x) \to 0$ for every $x \in X$. Is it true that $E$ determines weak-* convergence if and only if $E$ is nonmeager?
Q2a. If not, what are necessary and sufficient conditions for $E$ to determine weak-* convergence?
Clearly it is necessary that $E$ be dense. A rookie mistake is to guess that density is also sufficient, but this is false and it is easy to find counterexamples.
By a version of the uniform boundedness principle and the triangle inequality, every nonmeager subspace determines weak-* convergence (see below for a sketch). This is a very strong condition, since as discussed in this question, a nonmeager proper subspace of $X$ lacks the Baire property and so must be rather pathological, not a space we are likely to encounter in everyday life. For practical purposes, requiring that $E$ be nonmeager is essentially as strong as requiring that $E = X$.
I was wondering whether "nonmeager" is in fact necessary.
Suppose we could write $E = \bigcup_n E_n$ where $E_n$ are increasing nowhere dense subspaces. Since $E_n$ is nowhere dense, it is not dense, so by Hahn-Banach we may find $f_n \in X^*$ with $f_n(E_n) = 0$ and $\|f_n\| = n$. Then $f_n(x) \to 0$ for every $x \in E$, but $\{f_n\}$ is unbounded so by the uniform boundedness principle, there exists $x \in X$ with $\{f_n(x)\}$ unbounded. So such an $E$ does not determine weak-* convergence.
Therefore, if Q1a has an affirmative answer, so does Q2.
Footnote: Here's a sketch of the argument that nonmeager subspaces determine weak-* convergence. Suppose $E$ is nonmeager and $\{f_n\} \subset W^*$ with $f_n(x) \to 0$ for $x \in E$. Let $A_k = \bigcap_n \{x : |f_n(x)| \le K\}$. $A_k$ is closed, and since $\{f_n(x)\}$ is bounded for $x \in E$, we have $E \subset \bigcup_k A_k$. By Baire, some $A_k$ has nonempty interior, and a little rearrangement shows that $M := \sup_n \|f_n\| < \infty$, as in the usual uniform boundedness principle. Now since $E$ is nonmeager, it must be dense, since non-dense subspaces are nowhere dense. For any $x \in X$, fix $\epsilon$ and choose $x' \in E$ with $\|x-x'\| < \epsilon$. Then $|f_n(x)| \le |f_n(x')| + M \epsilon$; letting $n \to \infty$ we get $\limsup_n |f_n(x)| \le M \epsilon$, but $\epsilon$ was arbitrary so $f_n(x) \to 0$.
Best Answer
Update: Here is a ZFC (probably even ZF+DC) counterexample for Q1. It's from probability and kind of indirect, maybe someone will be able to find something shorter.
Let $X = C_0([0,1])$, the space of continuous functions $\omega$ on $[0,1]$ having $\omega(0)=0$. Fix $0 < \alpha < 1/2$ and let $E = C^{0,\alpha}([0,1]) \cap X$ be the subspace of $\alpha$-Hölder continuous functions. There are several ways to see that $E$ is meager in $X$. For instance, by Arzelà-Ascoli, the balls of $E$ under the Hölder norm are compact in $X$, hence $E$ is a countable union of compact sets, each of which is nowhere dense in $X$ by Riesz's lemma.
Or, the argument mentioned in the question: $E$ is Banach in its own norm, hence analytic in $X$, hence has the Baire property in $X$, hence must be meager.(That doesn't work because $E$ is not separable in its own norm, hence not Polish.)Now let $\mu$ be the Wiener measure on $X$. It is well known that $\mu(E) = 1$, since Brownian motion is almost surely $\alpha$-Hölder continuous for any $\alpha < 1/2$. On the other hand, suppose $F$ is any nowhere dense subspace of $X$, so that its closure is a proper closed subspace of $X$. By Hahn-Banach there is a nonzero continuous linear functional $f$ that vanishes on $F$. Now $\mu$ is a nondegenerate Gaussian measure on $X$, so $f$ has a nondegenerate one-dimensional Gaussian distribution under $\mu$; in particular, $\mu(F) \le \mu(\{f = 0\}) = 0$. Thus every nowhere dense subspace has measure zero, so by countable additivity, $E$ cannot be a countable union of such.
Previous answer. It is consistent with ZF+DC that the answer to Q1 is No.
Consider an example in which $E$ is separable Banach in a stronger norm $\|\cdot\|_E$; for instance, $X = L^2([0,1])$ and $E = C([0,1])$. Then $E$ is analytic in $X$, hence has the Baire property, hence must be meager. If $E$ can be written as a countable union of subspaces $E_n$ then by Baire category one of them must be nonmeager with respect to $\|\cdot\|_E$, hence doesn't have the BP in $E$. But Shelah proved it is consistent with ZF+DC that every subset of a Polish space has the BP.
In our specific example we can see explicitly that $C([0,1])$ is meager in $L^2([0,1])$: the sup-norm balls $B_n = \{f : \|f\|_\infty \le n\}$ are closed in $L^2$ with empty interior.
I don't see whether this helps us with Q2.