$\sup A = \inf B$ implies $\forall\varepsilon>0.\exists a\in A, b\in B. b-a<\varepsilon$ Let $A, B$ two sets such that $\sup A = \inf B$. Is it right that:
$$
\forall \varepsilon > 0. \exists a\in A, b\in B. b-a<\varepsilon \quad ?
$$
The question doesn't mention the sets are densed, but that was probably the intention.  
I think the claim is true.
for every $\varepsilon > 0$ we are able to choose $a\in A$ such that $\sup A - a < {\varepsilon \over 3}$ and $b\in B$ such that $b - \inf B < {\varepsilon \over 3}$.
And so, the distance is ${2 \over 3\varepsilon} < \varepsilon$.
We can choose $a, b$ as mentioned because those sets are densed.
Is that right? (Or am I being silly here? because the proof is almost trivial).

Yes, that's right. You can always make that choice, by definition of $\sup A$ (if that's not a maximum then you can choose elements at distance less that $\varepsilon$, else $supA-\varepsilon$ would be a smaller $\sup A$). 
FALSE( Denseness is needed if you have to choose an element with distance exactly $\varepsilon$ )
