Baire category in $\mathbb{R}$, by nested intervals with canonically chosen rational endpoints: a countable intersection of dense open sets is dense, so $\mathbb{R}$ is not a countable union of nowhere dense sets

Fix $x_0 \in \mathbb{R}$ and a real $\varepsilon_0 > 0$; by [L1] it suffices to produce a point of $\bigcap_n U_n$ lying in $N_{\varepsilon_0}(x_0)$, since $x_0$ and $\varepsilon_0$ are then arbitrary.

By [L4] fix a bijection $\beta : \mathbb{N} \to \mathbb{Q}$ and put $e := \iota \circ \beta$, where $\iota(q) = \hat q$, so that $e$ is a bijection from $\mathbb{N}$ onto $\mathbb{Q}_{\mathbb{R}}$.

Recall the terminology of the Given: a pair $(p,q)$ of elements of $\mathbb{Q}_{\mathbb{R}}$ is good when $p < q$, and $G$ is the set of good pairs.

Refinement claim. For every good $(p,q)$ and every $n \in \mathbb{N}$ there is a good $(p',q')$ with $[p',q'] \subseteq (p,q) \cap U_n$. To see it, note first that $(p,q)$ is nonempty, since [L4] supplies an element of $\mathbb{Q}_{\mathbb{R}}$ strictly between $p$ and $q$, and that $(p,q)$ is open by [L2]; fix $y_1 \in (p,q)$ and, by [L2], a real $\rho_1 > 0$ with $N_{\rho_1}(y_1) \subseteq (p,q)$. Since $U_n$ is dense, [A1] and [L1] give $y \in N_{\rho_1}(y_1) \cap U_n$, so $y \in (p,q) \cap U_n$, and that set is open by [A1], [L2] and [L3], so there is a real $\rho > 0$ with $N_\rho(y) \subseteq (p,q) \cap U_n$. By [L4] fix $p', q' \in \mathbb{Q}_{\mathbb{R}}$ with $y - \rho < p' < y < q' < y + \rho$. Then $p' < q'$, so $(p',q')$ is good, and every $t \in [p',q']$ satisfies $y - \rho < p' \le t \le q' < y + \rho$, whence $|t - y| < \rho$ and $t \in N_\rho(y)$; thus $[p',q'] \subseteq N_\rho(y) \subseteq (p,q) \cap U_n$.

Successor rule. For $(k, (p,q)) \in \mathbb{N} \times G$ let $m$ be the least natural for which some natural $j$ makes $(e(m), e(j))$ good with $[e(m), e(j)] \subseteq (p,q) \cap U_k$, and let $j$ be the least natural with that property for that $m$; put $T(k,(p,q)) := (\sigma(k), (e(m), e(j)))$. The set of eligible $m$ is nonempty by step 2.1 applied with $n = k$, since $e$ is onto $\mathbb{Q}_{\mathbb{R}}$ by step 1.2, so both minima exist by [L5] and $T : \mathbb{N} \times G \to \mathbb{N} \times G$ is a total function defined without any selection.

The recursion. By [L4] fix $p_0, q_0 \in \mathbb{Q}_{\mathbb{R}}$ with $x_0 - \varepsilon_0 < p_0 < x_0 < q_0 < x_0 + \varepsilon_0$; then $(p_0,q_0)$ is good and, as in step 2.1, $[p_0,q_0] \subseteq N_{\varepsilon_0}(x_0)$ by [L2]. Apply [L6] with $Y = \mathbb{N} \times G$, seed $(0,(p_0,q_0))$ and map $T$ to get $h : \mathbb{N} \to \mathbb{N} \times G$ with $h(0) = (0,(p_0,q_0))$ and $h(\sigma(k)) = T(h(k))$; an induction on $k$ shows that the first coordinate of $h(k)$ is $k$, so write $h(k) = (k,(p_k,q_k))$, every $(p_k,q_k)$ being good.

Write $I_k := [p_k, q_k]$, a nonempty closed bounded interval by [L2]. The rule of step 3.1 gives, for every $k \in \mathbb{N}$, that $I_{k+1} \subseteq (p_k,q_k) \cap U_k \subseteq I_k$; in particular the family $(I_k)$ is nested and $I_{k+1} \subseteq U_k$.

By [L7] applied to the nested family $(I_k)$ of nonempty closed bounded intervals, $\bigcap_{k} I_k \ne \varnothing$; fix $x$ in it.

For every $n \in \mathbb{N}$ one has $x \in I_{n+1} \subseteq U_n$ by steps 5.1 and 6.1, so $x \in \bigcap_n U_n$; and $x \in I_0 \subseteq N_{\varepsilon_0}(x_0)$ by steps 4.1 and 6.1. So $N_{\varepsilon_0}(x_0)$ meets $\bigcap_n U_n$.

Since $x_0 \in \mathbb{R}$ and the real $\varepsilon_0 > 0$ were arbitrary, every neighbourhood of every point of $\mathbb{R}$ meets $\bigcap_n U_n$, so that set is dense by [L1].

For the consequence, let $(A_n)$ be a sequence of nowhere dense sets and put $U_n := \mathbb{R} \setminus \overline{A_n}$, which is open by [L3] and [L8] and dense by [L8]; by step 8.1 the set $\bigcap_n U_n$ is dense, hence nonempty, and any $x$ in it lies outside every $\overline{A_n}$ and so outside every $A_n$, giving $x \notin \bigcup_n A_n$ and therefore $\bigcup_n A_n \ne \mathbb{R}$. By [L9] the same conclusion covers a union of an at most countable family of nowhere dense sets, so no meager set is all of $\mathbb{R}$.