Q: Deduce that if $\alpha$ and $\beta$ are two rational numbers so that for every prime $p$ we have $|\alpha-\beta|_p \leq 1$, and $|\alpha-\beta|_R < 1$, then $\alpha$ equals $\beta$.
In fact, there is a part (i) before this question. Part (i) is to show that for a rational number $\alpha$, prove $\alpha\in\mathbb{Z}_p$ for all primes $p$ if and only if $\alpha$ is an integer. And I finished this part (i guess i am correct...).
By the way, this question is a part (ii). It gives us a hint that the difference $\alpha-\beta$ must be an integer (by i), then $|\alpha-\beta|_R \leq 1$ forces an equality. What does it means...and how can I do this question? Thank you very much~~
Assuming part (i) you have mentioned, the answer to your question is straightforward: $\alpha - \beta$ is an integer whose (archimedean) absolute value is less than $1$, so it's $0$.
For a proof to part (i), here's a good first step: write $\alpha = a/b$ in lowest terms, and pick a prime $p$ dividing $b$, assuming $b \neq 1$.
(Of course, the comment by Lubin also gives a proof to part (ii), and his comment is probably worth your while thinking why it's true.)
(One more thing that I guess is worth nitpicking: usually people write $\mathbb{R}$ (\mathbb{R}) instead of $R$ for real numbers.)