prove/ disprove that every non empty subset of Q that is bounded above has a supremum in Q
??? You titled this "proof that an empty set is bounded"! This has nothing to do with the empty set!
Since that is false, a counter-example suffices. Let A= {3, 3.1, 3.14, 3.141, 3.1415, 3.14159, ...} where each number in the sequence is $\displaystyle \pi$ to one more decimal place. Since the real numbers are "complete", this statement is true for R- there is a supremum in R. If this set had a supremum in Q, it would have to be the same as the supremum in R. What is the supremum of this set (in R). Is that a rational number?