A nonempty perfect set in $\mathbb{R}$ which contains no rational number

When I am doing some measure theory exercises today, I suddenly think of this problem in chapter 2 of Rudin's Principles of Mathematical Analysis which I found difficult in my first year undergraduate study but seems easy to me now. I want to present the construction here.

We want to imitate the construction of the Cantor set so that we obtain a perfect set. Fix some closed interval $[a,b]$ with irrational endpoints and let $\{q_0, \;q_1, \;q_2, \dots\}$ be an enumeration of the rational points in $[a,b]$. We will remove all the $q_n$ from the interval $[a,b]$ in the process of constructing a decreasing sequence of subsets $S_n$ satisfying the properties: (i) $S_n$ is compact, (ii) $S_n$ has nonempty interior, (iii) the boundary of $S_n$, denoted by $\partial S_n$, consists of irrational numbers and (iv) $\partial S_n \subset \partial S_{n+1}$.

We do this: Set $S_0=[a,b]$. Then $S_0$ satisfies properties (i), (ii) and (iii). To construct $S_{n+1}$ from a given $S_n$ which satisfies assumption (i), (ii) and (iii), we find the least $k$ such that $q_k\in S_n$ (which exists by assumption (ii)). By assumption (iii), $q_k$ is not on the boundary, hence is in the interior of $S_n$. So we can find an open interval $(r,s)$ with irrational endpoints lying in the interior of $S_n$, i.e. $r, s \in {S_n}^{\circ} \backslash \mathbb{Q}$. Set $S_{n+1} = S_n \backslash (r,s)$. We see that $S_{n+1}$ is compact by assumption (i), $S_n$ has nonempty interior because some small interval $(r-\delta, r] \subseteq S_{n+1}$ and $\partial S_{n+1} = \partial S_n \cup \{r,s\}$ consists of irrational points. So assumptions (iii) and (iv) hold for $S_{n+1}$.

Let $S = \cap_{n=0}^\infty S_n$. We claim that $S$ is the desired set: $S$ is nonempty and closed as an intersection of nonempty compact sets. Since at least one rational number is removed from $S_n$ in each step, $q_n \notin S_{n+1}$ hence $S$ contains no rational number. Also, every point in $S$ is a limit point of $S$. To see this, notice that $\partial S_n \subseteq S$ for every $n$. If $x \in S$ is an irrational point which is not a right endpoint of any $S_n$, then for any given $\epsilon > 0$ we can find a rational $q \in [a,b]$ such that $x < q$ and $|x-q| < \epsilon$. Consider the interval $(r,s)$ which contains $q$ being removed in some step. We have $x \leqslant r < q$, and $x \neq r$ because $x$ is not a right endpoint, then $r \in \partial S_n \subseteq S$ for some $n$ and $|x-r| < \epsilon$. Note that by our construction a point cannot be a right endpoint and a left endpoint at the same time, so if $x$ happens to be a right endpoint we can take $q < x$ and consider $q < s < x$, so we have proved that every $x \in S $ is a limit point of $S$. Therefore $S$ is the desired perfect set.