Mentioned as an easy exercise in 'Topology', James R. Munkres, 2e, Pearson[page No:25]
If $A$ is an ordered set has the Least upper bound property iff it has the greatest lower bound property.
An ordered set $A$ is said to have the least upper bound property if every nonempty subset $A_0$ of $A$ that is bounded above has least upper bound. $A$ satisfies this property. Suppose $B_0$ is an arbitrary set having lower bound. We need to prove it has the greatest lower bound. Suppose $L\subset A$ is the set of all lower bounds of the set $B_0$. $$\exists b_0\in B_0, \forall x\in L, x\leq b_0. $$ So, $L$ is bounded above. Hence, It has least upper bound. Let $b$ be the least upper bound of $L$. $$ \forall x\in L, x\leq b. $$ If $\exists b'\in B_0, b'\leq b,$ which contradict the fact that $b$ is the least upper bound of $L$. Hence,$$b\leq y, \forall y\in B_0.$$ So, $b$ is the lower bound. No element of $y\in L$ satisfy $ b\leq y$. Hence $b$ is the greatest lower bound. If this proof is correct, I can use the similar argument for the converse. I request you to verify my proof.