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From my point of view I think it isn't but I'm not sure. Here is what I thought:

$(0,1) \bigcap \mathbb{Q}$ is basically $$ A = \{ \frac a b \in \mathbb{Q} \mid \frac a b \gt 0 \text{ and } \frac a b \lt 1 \}$$ and $\forall x \in A$ and $\forall R \gt 0$ there is a $q \in \mathbb{R} \setminus \mathbb{Q}$ such that $q \in \mathcal{B}_R(x)$ (i.e., open ball of center $x$ and radius $R$), so $\forall x \in A, \mathcal{B}_R(x) \nsubseteq \mathbb{Q}$ so $A \notin $ standard topology of $\mathbb{R}$. Conclusion, $A$ is not an open set in the standard topology of $\mathbb{R}$.

Is that correct ?

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    Yes that's correct. Great! – 2017-01-29
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    Oh for once I did it . Thank you ! – 2017-01-29
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    Also $q$ depends on $x$ and $R$. So for more clarity you can write $q(x,R).$ – 2017-01-29
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    Yes, you are right,there are so many different notation for things. – 2017-01-29
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    Your question (in the header) asks if it is an open set in *a* topology of $\Bbb R$; the answer to this is "yes": in the discrete topology on $\Bbb R$, all sets are open. I assume that what you intended was "is it open in the standard topology on $\Bbb R$?", in which case your answer is correct. – 2017-01-29
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    Yea , that was what I meant . We have not learned yet about discrete topology In my university,so I have no knowledge of that .I will edit the header for clarity, thank you ! – 2017-01-29

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