In Counterexamples in topology, when discussing example 82. Niemytzki's Tangent Disc Topology, the authors seemingly claim the following: Let $S_1 \subset S_2 \subset S_3 \subset \dotsc$ be a sequence of sets whose union are the irrational numbers, $$\bigcup_{i=1}^\infty S_i = \mathbb R \setminus \mathbb Q.$$ Then there is $n_0 \in \mathbb N$ such that $S_{n_0}$ is dense in some interval $(a,b)$. In particular, $S_{n_0}$ has some rational points boundary points.
Is that true? I thought the standard construction which shows that $\mathbb Q$ has measure $0$ might be able to produce a counterexample: Choose an enumeration of $\mathbb Q$, say $\varphi: \mathbb N \to \mathbb Q$. For $n \in \mathbb N$ take a union of intervals $$U_n = \bigcup_{k \in \mathbb N} \left(\varphi(k)-\frac{1}{2^{k+1}n}, \varphi(k)+\frac{1}{2^{k+1}n}\right),$$ so that $U_n$ is an open set of length $\frac{1}{n}$ containing $\mathbb Q$. This means the complement $$S_n = \mathbb R \setminus U_n.$$ consists of irrational numbers only. Also, $S_n$ is closed, so it doesn't have any rational boundary points. I'm just not sure when $\bigcup_n S_n = \mathbb R \setminus \mathbb Q$ holds. This should depend on the chosen enumeration $\varphi$. For example if the sequence $k \mapsto \varphi(2k)$ converges to an irrational number $x \in \mathbb R \setminus \mathbb Q$ faster than the interval lengths decreases, then $x$ is not in any of the $S_n$.
This follows almost immediately from the Baire category theorem. If no $S_n$ is dense in any interval, then they are all nowhere dense, as is the singleton $\{q\}$ for each $q\in\mathbb{Q}$. This gives a countable family of nowhere dense sets whose union is all of $\mathbb{R}$, contradicting the Baire category theorem.