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.
 A: 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!
A: The proof of existence of supremums that I saw relied on the completeness of the real numbers, and not much else.  Basically, we constructed a Cauchy (and thus convergent) sequence of real numbers that was always an upper bound of the set, and got infinitely close to the set.
Here's an outline:
Theorem. Let $U \subset \mathbb{R}$ be a bounded set of real numbers.  Then there exists $s \in \mathbb{R}$ such that for all $u \in U$, $s \geq u$, and for any $\varepsilon > 0$, $s - \varepsilon$ is not an upper bound of $U$.
Let $M > 0$ be the upper bound for $U$ (for example, if $U = (0,1)$, we could say $M = 1000000000$, and it would be fine).
Take $T_1 := M$, and $B_1$ to be some number that is not an upper bound for $U$.  
Now, we given $T_i$ and $B_i$, we will define $T_{i+1}$ and $B_{i+1}$ as follows:
We take the midpoint of $T_i$ and $B_i$ to be called $m_i$, and see if $m_i$ is an upper bound for $U$.  If it is, then we'll take $T_{i+1} := m_i$ and $B_{i+1} := B_i$.  We define $a_i$ to be $m_i$
If $m_i$ is not an upper bound for $U$, then we define $a_i := a_{i-1}$, and $T_{i+1} := T_i$, and $B_{i+1} := B_i$.
The distance between $T_i$ and $B_i$ halves for each iteration, so since $a_i$ is contained in the interval, it is squeezed into convergence.  Its limit $a$ must be an upper bound, and the claim (I will leave for you to play with) is that $a$ is the least upper bound of the set $U$. $\Box$
A: 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 $U$.   
