Given a partially ordered set $(X,\leq)$, is it true that $$ \forall P\subseteq X:P\neq\emptyset\text{ and }P\text{ is bounded below }\to P\text{ has a minimum}\\ \updownarrow\\ \forall P\subseteq X:P\neq\emptyset\text{ and }P\text{ is bounded above }\to P\text{ has a maximum} $$
If we replace minimum and maximum by infimum and supremum respectively, then a proof has been given here. However, in that proof, I don't see how one would infer $g\in P$ from $g\in B$.
On
The way I see it (and I could be wrong), the least upper bound property is a matter of the metric space and not the sets. In this excercise, matters of min and max are dependent upon the sets themselves. (As space is unspecified.) Can easily set up a discrete set such as {$1 - 1/n$} where all bounded below subsets have a min min but but bounded above subsets don't have a max.
On the other hand, had the metric space been specified we can find spaces (the integers for example) where the space would have this property. (Because the space is "superset" the deterimines all subsets.)
In the linearly ordered set $X = \{-\frac{1}{n}: n=1,2,\ldots\} \cup \{0\}$ (in the order inherited from the reals) we have that $X$ satisfies the upper condition (check this), but not the lower, because the left part has an upper bound $0$, but no maximum.
So the properties are not equivalent. Use $X = \{0\} \cup \{\frac{1}{n}: n=1,2,\ldots\}$ for the other implication, of course.