Doubt about continuity of a function

Question

Consider the function $f : \mathbb{R} \longrightarrow \mathbb{R}$ defined by

\begin{equation*} f(x)=\begin{cases} 1 + \frac{1}{q}, \quad &\text{if} \, x = \dfrac{p}{q} \in \mathbb{Q}. \\ 1, \quad &\text{otherwise}. \\ \end{cases} \end{equation*}

I need proof that the function is continuous in $\mathbb{R} \setminus \mathbb{Q}$ and discontinuous in $\mathbb{Q}$. Prove that f is discontinuous in $\mathbb{Q}$ is very easy. Let $\dfrac{p}{q} \in \mathbb{Q}$ and consider the sequence $\dfrac{p}{q} + \dfrac{1}{n}$. Then

$$\dfrac{p}{q} + \dfrac{1}{n} \longrightarrow \dfrac{p}{q} \quad \text{and} \quad f\bigg(\dfrac{p}{q} + \dfrac{1}{n} \bigg) \not\longrightarrow f\bigg(\dfrac{p}{q}\bigg).$$

My doubt is: How i proof that f is continuous in $\mathbb{R} \setminus \mathbb{Q}$?

I took an irrational number i and an arbitrary sequence of real numbers that converges to i and show that the function applied to this sequence converges to f(i)
I took an irrational number i and an arbitrary sequence of irrational numbers that converge to i and show that the function applied to this sequence converges to f(i) (is it a constant sequence and therefore trivial?)

Option 1 seems difficult to me and I couldn't get anywhere. Option 2 is very easy.

Note that if $x_n = \frac{p_n}{q_n}$ is a sequence of rational numbers converging to a irrational number, then we ultimately have $q_n \rightarrow \infty$ as $n\rightarrow\infty, see https://math.stackexchange.com/questions/1465812/sequence-of-rationals-with-an-irrational-limit-have-denominators-going-to-infini — stange, Oct 07 '23 at 22:26
In the definition of $f(x)$, presumably $f(x) = 1+1/q$ if $x \in \mathbb Q$ and $x=p/q$ where $p\in \mathbb Z, q\in \mathbb Z_{>0}$ and $p$ and $q$ are coprime? — krm2233, Oct 07 '23 at 22:55

Doubt about continuity of a function

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