If $A$ is an open subset of $(0,1),$ then $\chi_A$ is Riemann integrable on $[0,1].$
My attempt: Since $A$ is an open subset of $(0,1),$ we can write $A$ as countable union of disjoint open intervals in $(0,1).$ So, we have that $$A=\bigcup_{n=1}^{\infty}(a_n,b_n),$$ where $a_i,b_i \in (0,1), a_i<b_i,$ for all $i \in \mathbb{N}$ with $(a_m,b_m)\cap(a_n,b_n)=\varnothing$ whenever $m\ne n.$
Now, we have that $$\chi_A=\begin{cases} 1 & \text{ if } x\in A \\ 0 & \text{ if } x\notin A \\ \end{cases}$$
Observation: The set of points of discontinuity of the function $\chi_A$ is precisely given by $D=\{a_i,b_i:i\in \mathbb{N}\}.$
Theorem: Suppose $a<b$ and $f:[a,b] \rightarrow \mathbb{R}$ is a bounded function. Then $f$ is Riemann integrable if and only if $$|\{x \in [a,b]:f \text{ is not continuous at }x \}|=0,$$ where $|\cdot|$ is the outer measure.
We can clearly see that $|D|=0,$ since D is countable.
Hence, $\chi_A$ is Riemann integrable, from the theorem stated.
Is there any logical flaw in my proof?
Please let me know. Any comments on the proof is highly appreciated. Thanks!
Your observation regarding points of dis-continuity is not correct. For a simple example, consider the characteristic function of $\bigcup_n (\frac 1 {2n}, \frac 1 {2n-1})$. This function is not continuous at $0$. You can have more complicated situations, but I just gave a simlple example.
What is true is $D=\partial A$.
In fact, what you are trying to prove is false. If $C$ is a fat Cantor set in $[0,1]$ and $A$ is its complement, then $\partial A=C$ which has positive measure, so characteristic function of $A$ is not Riemann integrable.