Prove a Continuous Distribution Function is Uniformly Continuous 
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).
 A: Given a bounded monotone function $f:\mathbb{R}\to\mathbb{R}$, $\lim_{x\to-\infty} f(x)=a$ and $\lim_{x\to\infty} f(x)=b$ exists, by monotone convergence theorem.

Given a continuous function $f:\mathbb{R}\to\mathbb{R}$ s.t. $\lim_{x\to-\infty} f(x)=a$ and $\lim_{x\to\infty} f(x)=b$ exists, then $f$ is uniformly continuous.

Proof
$\forall\epsilon>0$:
$\qquad$ Let $e=\epsilon/3$.
$\qquad$ $\exists M_1\in\mathbb{R}^+$ s.t. $f((-\infty,-M_1))\subseteq(a-e,a+e)$.
$\qquad$ $\exists M_2\in\mathbb{R}^+$ s.t. $f((M_2,\infty))\subseteq(b-e,b+e)$.
$\qquad$ Let $M=\max(M_1+1,M_2+1)$.
$\qquad$ $f$ is continuous on $[-M,M]$, and so it is uniformly continuous.
$\qquad$ $\exists d>0$ s.t. $\forall x,y\in [-M,M], |x-y|<d\implies |f(x)-f(y)|<e$.
$\qquad$ $\forall x,y\in\mathbb{R}$:
$\qquad\qquad$ If $|x-y|<d$:
$\qquad\qquad\qquad$ If $x,y\in [-M,M]$, then $|f(x)-f(y)|<e$.
$\qquad\qquad\qquad$ If $x,y\in (-\infty,-M)$, then $|f(x)-f(y)|\le|f(x)-a|+|a-f(y)|<2e$.
$\qquad\qquad\qquad$ If $x\in (-\infty,-M),y\in [-M,M]$, then:
$\qquad\qquad\qquad\qquad$ $|f(x)-f(y)|\le|f(x)-a|+|a-f(M)|+|f(M)-f(y)|<3e$.
$\qquad\qquad\qquad$ All the cases are similar to the above, and so $|f(x)-f(y)|<3e\le\epsilon$.
A: $f$ is uniformly continuous on $\mathbb{R}\iff\forall\epsilon>0,\exists\delta>0$ s.t. $\forall x,y\in\mathbb{R},(|x-y|<\delta\implies|f(x)-f(y)|<\epsilon)$
If not, then:
$\qquad$ $\exists\epsilon>0$ s.t. $\forall\delta>0,\exists x,y\in\mathbb{R}$ s.t. $|x-y|<\delta$ and $|f(x)-f(y)|\ge\epsilon$.
$\qquad$ Substitute $\delta=1/n$ sequentially to obtain the sequences
$\qquad$ $(x_n),(y_n)$ which satisfy $|x_n-y_n|\to 0$ and $|f(x_n)-f(y_n)|\ge\epsilon$.
$\qquad$ If $(x_n)$ is not bounded above:
$\qquad\qquad$ Take a subsequence which $\to\infty$.
$\qquad\qquad$ The corresponding subsequence for $(y_n)$ also $\to\infty$.
$\qquad\qquad$ This contradicts the limit at infinity.
$\qquad$ So, $(x_n)$ is bounded above.
$\qquad$ Similarly, $(x_n)$ is bounded below.
$\qquad$ Take a convergent subsequence of $(x_n)$ which $\to t$.
$\qquad$ The corresponding subsequence for $(y_n)$ also $\to t$.
$\qquad$ This contradicts the continuity at $t$.
A: Not really an answer, but I can't comment.  
You won't be able to prove that $F$ is Lipschitz continuous.  Consider the Cantor function to see an example of a CDF which is not Lipschitz continuous. http://en.wikipedia.org/wiki/Cantor_function#Properties
A: Thanks to all who answered here. I thought of a solution, and whether it is correct or not, I leave it for you to judge.

$F$ is increasing, continuous, so consider any sequence $\{x_n\}\subset[0,1]$ such that $x_n\to0$.
Then, we will get a positive divergent sequence $\{L_n\}$ such that $[x_n,1-x_n]\subset F([-L_n,L_n])$. So $F(L_n)\geq1-x_n$ and $F(-L_n)\leq x_n$.
Thus, $0\leq \lim_{n\to\infty} F(-L_n)\leq\lim_{n\to\infty} x_n=0$ giving $F(\lim_{n\to\infty}-L_n)=0$ by Sandwich Theorem and continuity of $F$. Analogously it follows that $F(\lim_{n\to\infty}L_n)=1$.
Now we know that $F$ being continuous over $\mathbb R$, $F$ is uniformly continuous in the closed and bounded interval $[-L_n,L_n]$ for all $n\in\mathbb N$.
But, $$ [-L_n,L_n]\subset[-L_{n+1},L_{n+1}],and\\\cup_{n\in \mathbb N}[-L_n,L_n]=\mathbb R.$$
$F$ is uniformly continuous in each $[-L_n,L_n]$ so $F$ is uniformly continuous in whole of $\mathbb R$

