functions - Let $f(x) = 1$ for rational numbers $x$, and $f(x)=0$ for irrational numbers. Show $f$ is discontinuous at every $x$ in $mathbb{R}$

I am working on this proof, and wanted someone to check it and to help me understand what is happening in case (ii). The proof:

Let $f(x) = 1$ for rational numbers $x$, and $f(x)=0$ for irrational numbers. Show $f$ is discontinuous at every $x$ in $\mathbb{R}$.

We will consider two cases.

(i) $x \in \mathbb{Q}$

Consider the sequence $x_n = x + \frac{\sqrt{2}}{n}$. We have that $(x_n) \rightarrow x$, yet $x_n$ is irrational $\forall \ n$.
$\Rightarrow x_n \ \in \ \mathbb{Q}$

$\Rightarrow x_n - x \in \ \mathbb{Q}$.

$\Rightarrow n(x_n-x) = \sqrt{2} \in \mathbb{Q}$.$\ \rightarrow \leftarrow$

This is a contradiction, therefore we have that $f(x_n) = 0 \ \forall \ n \Rightarrow \lim f(x_n) =0 \neq 1 =f(x)$.

Therefore, $f(x)$ is not continuous at any $x \in \mathbb{Q}$.

(ii) $x \not\in \mathbb{Q}$

Given the density of the rationals, there exists a subsequence of rational numbers that must converge to $x$. Call this subsequence $(x_{n_r})$. Therefore, we have that:

$f(x_{n_r})=1 \ \forall \ n \Rightarrow \lim f(x_{n_r}) = 1 \neq 0 = f(x).$

$\therefore \ f(x)$ is not continuous at any $x \in \mathbb{Q}$.

By cases (i) and (ii), we have that $f(x)$ is not continuous at any $x \in \mathbb{R}$.

For case (i), I used the hint in the back of my text book to generate a sequence I could work with. Part (ii) is almost identical to the solution in my text- I don't fully understand why we are using the denseness of the rationals, rather than just working with the sequence made in case (i).

Answer

When $x$ is irrational, the sequence defined in case (i) might not consist of only rationals. For example, if $x = \sqrt 2$, then $x + \frac {\sqrt 2} n = \frac {n+1} n {\sqrt 2}$ is irrational. (It will converge to $x$, but it doesn't accomplish what's needed.)

For (ii), you need a sequence of rationals converging to the irrational $x$. In theory, we already know one: consider the decimal expansion of $x$. When $x$ is irrational, the sequence is necessarily infinite, doesn't eventually repeat itself forever. Suppose $$x = m + 0.d_1 d_2 \dotso$$ where $m$ is an integer. Let
$$x_n = m + 0. d_1 \dotso d_n$$
In other words, $x_n$ is the decimal representation of $x$ cut off at the $n^{th}$ digit after the decimal point ($x_0 = m$). Then every $x_n$ is rational, and:

$$lim_{n \to \infty} x_n = x$$