Discontinuity of Dirichlet function

Define $$f(x)=\begin{cases}1, & \text{if }x\in\mathbb{Q}, \\0, & \text{if }x\in\mathbb{R}\setminus\mathbb{Q}.\end{cases}$$Then $f$ has a discontinuity of the second kind at every point $x$, since neither $f(x+)$ nor $f(x-)$ exists.

Proof: We'll consider only for $f(x+)$.

Case 1. If $x_0\in \mathbb{Q}$ then we can take $t_n=x_0+\frac{1}{n}$ at that $t_n\to x_0,t_n>x_0$ and $t_n\in \mathbb{Q}$. Hence $f(t_n)=1\to 1$ as $n\to \infty$.

Also we can take $t_n=x_0+\dfrac{\sqrt{2}}{n}$ at that $t_n\to x_0,t_n>x_0$ and $t_n\in \mathbb{R}\setminus\mathbb{Q}$. Hence $f(t_n)=0\to 0$ as $n\to \infty$

Case 2. For $x_0\in\mathbb{R}\setminus\mathbb{Q}$ we apply a similar argument.

We can take $t_n=x_0+\dfrac{1}{n}$ and in this case $f(t_n)\to 0$. Taking $t_n\in \mathbb{Q}$ such that $x_0<t_n<x_0+\dfrac{1}{n}$ we get $f(t_n)\to 1.$

Hence $f(x_0+)$ does not exists at any point $x_0\in \mathbb{R}$. Also $f(x_0-)$ does not exists at any point $x_0\in \mathbb{R}$.

Hence Dirichlet function has discontinuity of the second kind at every point of $\mathbb{R}^1$.

Is my proof true?