Meager subspaces of a Banach space and weak-* convergence

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.