Bounded sets do have a least upper bound I think this is very fundamental property of real numbers, as it allows us to discuss about limits. I do not know, how the limits are playing an important role to complete the following one. Please explain.

" Any non-empty set of real numbers which is bounded above has a supremum" 

This is available in standard books. But, how to complete the proof of the cited above statment by introducing LIMITS, I don't know. Advanced thanks to MSE members, who can help me.
The fact that a nonempty set of real numbers bounded above has a supremum is often taken as an axiom in analysis courses. One can construct the set of real numbers from the set of rational numbers via Dedekind cuts, and then prove this as a theorem instead. This is done for example in Rudin's Principles of Mathematical Analysis, and the proof does not have anything to do with limits at all!
Let $U\subset\mathbb{R}$ be bounded above. Then there exists $x'\in\mathbb{R}$ such that $u\leq x$, for all $u\in U$. Therefore, $x'$ is the upper bound of $U$. Let, $X=\{x\in \mathbb{R}:u\leq x\,for\,u\in U\}$ be the set of all upper bounds of $U$. Then obviously $x'\in X$ and $x^*=\min X$ is the least upper bound of $U$. As $u\leq x^*$ for all $u\in U$ and $x^*\leq x$, for all upper bounds $x$ of $U$.