Proving $U_{\varepsilon}(a)\cap V_{\varepsilon}(b)=\emptyset$ I believe I have the right idea but having trouble formulating a mathematical argument for this proof. 

Suppose $a,b\in \mathbb{R}, a\neq b$.  Show there exists $\varepsilon$-neighborhoods $U_{\varepsilon}(a)$ and $V_{\varepsilon}(b)$ such that $U\cap V=\emptyset$

So i figure I start out saying, let $|b-a|>2\varepsilon$.  This establishes that the distance between $a$ and $b$ is more than the necessary $2\varepsilon$ distance.  Thus I need to find a number $x\in\mathbb{R}$, $a<x<b$ where $x\notin U, V$.  I'm supposing this number is $\frac{b-a}{2}$.  
But now how do I show that $\frac{b-a}{2}\notin (a-\varepsilon,a+\varepsilon)$ and $\frac{b-a}{2}\notin (b-\varepsilon,b+\varepsilon)?$
 A: Imagine this geometrically. Say there are two points $a, b$ on the line. Find the distance between the points and halve it. If you draw two open balls (circles such that it does not contain points on its perimeter) centered at each point with radius equal to half the distance between them then will they intersect??
Let $\left|{a - b}\right| = d \gt 0$ since $a \neq b$. Let $U_{\epsilon}(a) = \left|{x - a}\right| \lt \frac d 2$ and $V_{\epsilon}(b) = \left|{x - b}\right| \lt \frac d 2$ where $\epsilon = \frac d 2$ This is exactly what you suggested btw.
Now let's show that they are disjoint. 
Say $y \in U_{\epsilon}(a)$ and $y \in V_{\epsilon}(b)$
Since, $y \in U_{\epsilon}(a) \implies \left|{y - a}\right| \lt \frac d 2\implies \left|{(a - b) - (y - b)}\right| \lt \frac d 2 \implies \left|{a - b}\right| - \left|{y - b}\right| \lt \frac d 2 \implies d - \left|{y - b}\right| \lt \frac d 2 \implies \frac d 2 \lt \left|{y - b}\right| \implies y \not \in V_{\epsilon}(b)$ leading to a contradiction. 
$\implies V_{\epsilon}(b) \cap U_{\epsilon}(a) = \emptyset$
Your logic is false, since finding just one number which is not in both $U$ and $V$ does not prove that they are disjoint. You need to prove there are no common points in the two neighbourhoods. 
A: $$
\left|b-\frac{b-a}2\right|=\left|\frac{2b-b+a}2\right|=\left|\frac{b+a}2\right|>\frac{2\varepsilon}{2}=\varepsilon.
$$
Therefore $\frac{b-a}2$ doesn't live in $(b-\varepsilon,b+\varepsilon)$, since it's too far away. Similar argument for the interval around $a$. 
See below for a more thorough answer, though.
A: Sounds to me as if your main difficulty is with the logic rather than the algebra/analysis.  This is very common in beginning analysis proofs and it's worth spending some time on it.  So here are some hints.
First your statement "This establishes that the distance between $a$ and $b$ is more than the necessary $2\varepsilon$ distance".  This is not exactly wrong but it sounds as if you are trying to prove something about a pre-specified $\varepsilon$.  In fact, you have to choose an $\varepsilon$, and this idea is your basis for choosing it.  So you could say something like "choose some positive $\varepsilon<\frac{1}{2}|b-a|\,$".
Next, you say that you have to find some $x$ such that $x\notin U$ and $x\notin V$.  But this does not mean $U\cap V=\varnothing$.  In fact, you have to prove that no $x$ is in both $U$ and $V$.  Hint.  Assume $x\in U$ and $x\in V$, and derive a contradiction, using the value of $\varepsilon$ you chose earlier.
A: Let be $\epsilon = \frac{|b-a|}{4}$. Take balls centered on $a$ and $b$ with radium $\epsilon$.
