Let $F$ be the distribution function for a random variable $X$ and it is given that $F$ is continuous over the entire real line. Prove that $F$ is uniformly continuous over the real line.
My approach was something like this:
Take any two points $x$ and $y$ in $\mathbb R$. If $|x-y|>1$ then trivially $|F(x)-F(y)|\leq1<|x-y|$ hence $F$ is Lipschitz whenever $|x-y|>1$. So for all such points $x,y$ where $|x-y|>1$, $F$ is uniformly continuous.
However, I am stumped in the case where $|x-y|<1$. How do we finc a $\delta$ such that $|x-y|<\delta\implies|F(x)-F(y)|<\epsilon$, given any $\epsilon>0$?
Please give me some hint(s).