I know that there alredy exists numerous proof on this issue, for example https://math.stackexchange.com/a/1427790/579544, I have tried to prove the Riemann-integrability of Thomae function $f:[0,1]\to\mathbb{R}$,
$$f(x)=\begin{cases} 0 \text{ ; when } x \text{ is irrational} \\\frac 1 q \text{ ; for } x=\frac p q \text{ irreducible fraction}.\end{cases}$$ by means of a finite open cover of $[0,1]$.
However, I was wondering if the following step is legit:
We know that $f$ is continuous at every irrational point $r\in [0,1]$. So for an arbitrary $\epsilon>0$ we define $U_{\delta_r}(r):=\{x\in[0,1]\mid |x-r|<\delta_r\}$ such that $|f(x)-f(r)|<\epsilon$. As $[0,1]$ is compact every open cover has a finite open cover of $[0,1]$ so I can find fintely-many $U_{\delta_r}(r)$ such that $\bigcup\limits_{i=1}^N U_{\delta_{r_i}}(r_i)\supseteq [0,1]$.
If it is not allowed, is it possible to show the Riemann integrability in a different way as in the link above without the notion of measure?
EDIT:
I think my proposed idea is wrong and not very helpful, because $[0,1]$ is not a subset of $\bigcup\limits_{i=1}^N U_{\delta_{r_i}}(r_i)$. If I simply set $\epsilon:=\frac{1}{4}$ then we see that the point $\frac{1}{2}$ will never be an element of $\bigcup\limits_{i=1}^N U_{\delta_{r_i}}(r_i)$. So, if I am not able to make the $\epsilon$ small enough then this approach doesn't seem very promising at all when it comes to Upper/Lower-Riemann/Darboux-sums.
