Proposition: Let $L$ be a lattice in which every subset with an upper bound has a least upper bound. Then every subset with a lower bound has a greatest lower bound.
Attempt
Definition: A lattice, $L$, is a partially ordered set where given any two elements of $L$, $a$ and $b$, the set $\{ a,b \}$ has a least upper bound and a greatest lower bound.
Denote a subset of $L$ by $S$. If $S$ has an upper bound, then $S$ has a least upper bound, or more precisely, given $u \in L$, $\forall s \in S$, $s \le u$ and given any upper bound, $v$, of $S$, $u \le v$.
Suppose $S$ has a lower bound, $w$. Then $\forall s \in S$, $w \le s$.
My question
From here, I think I need to rely on the definition of a lattice. However, the definition only applies to a two element subset, where the proposition provides any subset.
Two candidates I thought that may lead to progression is the transitivity axiom for a partially ordered set, or a previous proposition I provided that any chain is a lattice. Presently, I am unsure on how to figure these in.
I would appreciate some assistance on what I need to do in order to complete this proof.
Source: Kaplansky, I. (1972). Set Theory and Metric Spaces.
Based upon the comments received, I think I figured it out.
Suppose that $S$ is has a lower bound, and denote the set of lower bounds by $B$. Since $B$ consists values that for each $b \in B$, $b \le s$, $\forall s \in S$. Thus $S$ must consist of upper bounds of $B$, and thus $B$ is bounded above. Therefore by the proposition $B$ has a least upper bound which corresponds the the greatest lower bound of $S$.