Two sets such that $\sup(A) = \inf(B)$, there are elements within a distance of $\epsilon$ 
Given two bounded, nonempty sets $A$ and $B$, such that $\sup(A) = \inf(B)$.
$(a)$ Show that $A \cap B$ contains at most $1$ element.

Suppose that two distinct $a, b \in A \cap B $, we then have:
$$ a \leq \sup(A) = \inf(B)$$
since $a \in  A$, and also since  $b \in B$ we can say that that:
$$ a \leq \sup(A) = \inf(B) \leq b \rightarrow a \leq b$$
We now make the same statement, but reverse the roles of $a$ and $b$, since $b \in  A$, and also since  $a \in B$
$$ b \leq \sup(A) = \inf(B) \leq a \rightarrow b \leq a$$
We conclude that $a=b$. $\square$
I think this proof is fine so far.

(b) Show that:
$$ \forall \epsilon >0: \space{  } \exists a \in A, b \in B: \space |a-b|< \epsilon $$

I think the point is to use the first exercise, 
1 element so suppose we have $c \in A \cap B$ then also $c \in A,B$ , we then have that $|c-c|=0<\epsilon$
No elements For the second case we would need to prove that whenever $A \cap B = \emptyset$, we can still find such an $a$ and $b$. Maybe we could somehow use the fact about the supremum and infimum, but I am not sure how to use this. We know that $a \leq \sup(A)=\inf(B) \leq b$, this just tells us how they are ordered.
 A: I think it's unnecessarily complicated to use the first question. It could be useful to use the following property of $c=\sup(A)$. For every $\varepsilon>0$ there exists $a\in A$ such that 
$$c-\varepsilon<a$$
(otherwise $c$ wouldn't be the lowest upper bound!). Similarly if $d=\inf(B)$ then for every $\varepsilon>0$ there exists $b\in B$ such that
$$b<d+\varepsilon$$
Can you finish it from there?
A: You can use the first statement if you wish. So you have two cases.
Case 1 A$\cap$B = {a} You'll want to show that a = inf A = sup B. Then you can prove your claim using definitions.
Case 2 A$\cap$ B = $\emptyset$ For elements to be within distance $\epsilon$ of each other, you can show they are distance $\frac{\epsilon}{2}$ away from s = inf A = sup B. This can also be done with definitions of sup and inf.
What @Olivier Moschetta has written is also correct. You'll notice you'll use the same arguments in both cases.
A: I don't think the first statement is necessary for the second.  
But the definition (one way or another) of $\sup A$ is that if $w < \sup A$ then $w$ is not upper bound.  So if $w < \sup A$ there exists an $a\in A$ so that $w < a \le \sup A$.  
Likewise if $v > \inf B$ then $v$ is not a lower bound of $B$ and so there is a $b \in B$ so so that $\inf B \le b < v$.
If we let $w = \sup A - \frac 12\epsilon$ and $v = \inf B + \frac 12 \epsilon $ we get that there are elements $a,b$ so that:
$w = \sup A - \frac 12\epsilon < a \le \sup A = \sup B \le b < \sup B + \frac 12 \epsilon$.
And so $|b-a| = b- a < (\inf B + \frac 12 \epsilon) - (\sup - \frac 12\epsilon = \epsilon$.
Now it could be that $a = b = \sup A=\inf B$.  Or it might not.  If $A\cap B = \{z\}$ where $z = \inf B=\sup A$ that's a perfectly good example that works.  But it need not be the only.  If $A\cap B =\emptyset$ then we know that there is another example.  But none of this is necessary
A: Let $c=\sup(A)$ and $d=\inf(B)$
For every $\varepsilon>0$ there exists $a\in A$ such that 
$$c-\frac{\varepsilon}{2}<a \rightarrow c-a <\frac{\varepsilon}{2} $$
Similarly,  for every $\varepsilon>0$ there exists $b\in B$ such that
$$b<d+ \frac{\varepsilon}{2} \rightarrow  b-d < \frac{\varepsilon}{2}$$
Also notice that $c=d$, so we have that 
$$b-c < \frac{\varepsilon}{2}$$
 Great, now notice that by the triangle inequality:
$$ |a-b|=|a-c + c-b| \leq |a-c| + |c-b|=|c-a|+|b-c|<\frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon   $$
Booya!
