In the proof above from Folland, I see $\mathcal{N}\subseteq\cup E_n$ where $\mathcal{N}$ is not meager. $\mathcal{N}=\cup_n ( E_n\cap \mathcal{N})$ is not countable union of nowhere dense sets. But from this $E_n\cap \mathcal{N}$ is not nowhere dense for some $n$ and contais an interior point from which $B(x_o,r)\subseteq E_n\cap \mathcal{N} \subseteq E_n $ and because $E_n$ has all of its limit points, $\overline{B(x_o,r)}\subseteq \overline{E_n}=E_n$.
If we are under the hypothesis of $(b)$, clearly we would have $\cup E_n=\mathcal{X}$. Because $\mathcal{X}$ is Banach, Baire Category Theorem shows it should not be decomposable into countable union of nowhere dense sets. So some $E_n$ is not nowhere dense and contais an interior point from which $B(x_o,r)\subseteq E_n $ and because $E_n$ has all of its limit points, $\overline{B(x_o,r)}\subseteq \overline{E_n}=E_n$. The rest of the proof is identical. So we can prove $(b)$ using Baire Category Theorem and I think this is what Folland remarks at the end of the proof.
So I understand we may modify the argument in (a) using Baire to yield (b) in the highlighted section. Is this corrrect?
$\mathcal{X}$ is nonmeager directly from Baire so it is also straightforward to prove it directly from $(a)$.
Yes, what you have stated is correct. The main idea of that proof to obtain part (b) is to apply the argument in part (a) in the case where $\mathcal{N} = \mathcal{X}$, which can be done as $\mathcal{X}$ is a Banach space and is consequently nonmeager by the Baire category theorem.