"Between every two rational numbers there exist infinite irrational numbers and between every two irrational numbers there exist infinite rational numbers.
Is this statement correct? If it is, then doesn't it contradict Thomae's function continuity at irrational numbers?
On
The statement is correct and it does not contradict the fact that Thomae's function is continuous precisely at the irrationals.
On
"Between every two rational numbers there exist infinite irrational numbers and between every two irrational numbers there exist infinite rational numbers."
This statement is true.
We know that, between any two rational numbers $\frac{a}{b}$ and $\frac{c}{d}$ with $\frac{a}{b}<\frac{c}{d}$ we can find a third rational number. One way would be to define the third rational as $$\frac{a}{b}+\frac{1}{2}\left( \frac{c}{d} - \frac{a}{b} \right) = \frac{bc+ad}{2bd}.$$ This process could be repeated to generate an infinite quantity of rational numbers between $\frac{a}{b}$ and $\frac{c}{d}$.
We can also find an irrational between any two rationals. Again using $\frac{a}{b}$ and $\frac{c}{d}$, establish a common denominator, leaving us with $\frac{ad}{bd}$ and $\frac{cb}{bd}$. Since $a$, $b$, $c$, and $d$ are integers and $ad<cb$, we know that $ad+1 \leq cb$. Let $q = \frac{\sqrt{2}}{2} <1$. It follows that $$\frac{ad}{bd} < \frac{ad+q}{bd} < \frac{cb}{bd}.$$ Thus, we can find an irrational between any two rational numbers.
Combining these two facts, we see that the first half of the statement, "between every two rational numbers there exist infinite irrational numbers", is true. The second half follows by very similar logic.
Many proofs of this fact rely on density of the rationals and irrationals in the reals. In this context, a set $S$ of numbers is dense in $\mathbb{R}$ if, for every interval $I=(a,b)$ where $a<b$, $I$ contains an element of $S$. You will run into this if you do any further reading on the subject.
As for your question about Thomae's function, your Wikipedia link already includes a clearly written informal proof of continuity at the irrationals. I don't know your current knowledge level, so if the following is too simple, please forgive me.
In order to understand the informal proof in your link, you will need to know that $\lceil x \rceil$ and $\lfloor x \rfloor$ refer to the ceiling and floor functions. You will also need the epsilon-delta definition of a limit. After that, the informal proof is quite approachable.
Between every two distinct rational/irrational numbers there are infinitely many irrational/rational numbers [respectively] - this is true.
But this does not contradict the continuity of Thomae's function at the irrational points. At an irrational number $r$ the value of the function is $0$. At nearby rational numbers the value is non-zero, but by restricting the interval $(r-\delta,r+\delta)$ by making $\delta$ sufficiently small, we can make sure that the only rationals in the interval have denominator $q\gt \frac 1{\epsilon}$ so that the value of the function at each of the infinitely many rational points in the interval is as small as we choose.