We start by assuming that we have a set $A$, that has the least upper bound property. Let's consider some arbitrary $A_0 \ \subset \ A$. Let $S_0 \ \subset \ A$ be the set of lower bounds for $A_0$. We know that $S_0$ will have a least upper bound (a supremum), by the least upper bound property. We let this least upper bound be denoted by $s$. If $s \ \in \ S_0$, then we're essentially done, as for $x \ \in \ S_0$, then $x \ \leq \ s$ (by definition of a least upper bound). If we assert that $s \ \notin \ S_0$, meaning that $s$ is not in the set of lower bounds for $A_0$, then it would follow that there exists some $x \ \in \ A_0$ such that $x \ < \ s$. This contradicts the fact that $s$ is a least upper bound for $S_0$, as $x$ is also an upper bound and clearly less than $s$. This implies that in order for $s$ to be a least upper bound of $S_0$, it must follow that $s \ \in \ S_0$, meaning that $S_0$ has a largest element, and the greatest lower bound property is satisfied.