Proof that the Dirichlet function is discontinuous

Let's have a look from the first point. Assume that the function $D(x)$ has limit $c$ at some point say $x_0$. Then if we choose $\epsilon=1/2$ then we have a number $\delta$ such that $$0<|x-x_0|<\delta\to|D(x)-c|<1/2$$ You do know as you point that every deleted neighborhood $0<|x-x_0|<\delta$ contains a rational point say $x_1$ and another point which is irrational say $x_2$. So regarding to what we have achieved, $$|D(x_1)-c|=|1-c|<1/2,~~~|D(x_2)-c|=|0-c|<1/2$$ and so we have reached to point of meeting a nice contradiction: $$1=|1-c+c|\leq ???<1/2+1/2=1$$

There always exist two distinct points: one is rational and the other is irrational in the interval $(x_0-\delta, x_0+\delta)$. By the definition of Dirichlet function, we can get the conclusion.

$\forall x\in \mathbb{R}, \space \forall \delta \gt 0, \exists x_{1}\in\mathbb{Q},x_{2}\in\mathbb{R}-\mathbb{Q}$, s.t. $x_{1},x_{2} \in (x-\delta, x+\delta)$.

This is because the density of rational number and irrational number in real line. I guess you should learn more about the construction of real number field.