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?

Yes, your proof is correct.
To summarize, the key point is to "construct", or show the existence of, if one wishes, two decreasing or increasing convergent sequences $(x_n)$ and $(y_n)$ such that
they both converge to the same given point,
but $(f(x_n))$ and $(f(y_n))$ have different limits.
You implicitly use the following two facts in your proof:

*

*(1) the sum of a rational and an irrational is irrational;

*(2) the set of rational numbers is dense in the real line.

You may also use the fact that the set of irrational numbers is dense in the real line together with (2) to give a proof.
