Question: Can anybody show me the following Proof?

Prove that if $X\_n$ for $n=1,2,\ldots,\infty$ is a real sequence that is uniformly distributed modulo $1$, and if $Y\_n$, $n=1,2,\ldots,\infty$ is a real sequence such that $\lim\_{n\to\infty} X\_n – Y\_n = \alpha$ then $Y\_n$ is uniformly distributed modulo $1$.

Remark: This includes the situation where $X\_n – Y\_n = \varepsilon$ tends to zero as $n$ tends to infinity.

Hint: Prove for zero case first. Show that $X\_n$ and $Y\_n$ are close.

Answer: 
Hint: for any $b \in [0,1]$, consider how many $Y\_n$ for $1 \le n \le N$ are in $[0,b]$ modulo $1$. If $|Y\_n - X\_n| < \epsilon$, what interval for $X\_n$ would guarantee $Y\_n$ is in this interval? What interval must $X\_n$ be in if $Y\_n$ is in this interval?
As an alternative to Robert's answer, it can also be done with the Weyl criterion. Let $X\_n-Y\_n=\alpha+b\_n$ with $b\_n\to0$. Then $$e^{2\pi ihY\_n}=e^{2\pi ihX\_n}e^{-2\pi ih\alpha}e^{-2\pi ihb\_n}$$ The second term on the right is a constant, so it can be pulled out of any sum. The third term, you should show it can be written as $1+c\_n$ with $c\_n\to0$.