9 Theorem A set $X$ is order-complete relative to an ordering if and only if each non-void subset which has a lower-bound has an infimum.
Suppose that $X$ is order-complete and $A$ is non-void subset which has a lower bound. Let $B$ be the set of all lower bounds of $A\,.$ Then $B$ is non-void and surely every member of the non-void set $A$ is an upper bound for $B\,.$ Hence, $B$ has a least upper bound, say, $b\,.$ ...
This is excerpted from General Topology's Orderings by John Kelley; here he proves the theorem that a set is order-complete iff all its non-void subsets which have lower bounds have infimum.
I did stumble a little and couldn't comprehend the statement "Hence, $B$ has a least upper bound $b\,.$"
Is it a trivial conclusion from the fact that all elements included $A$ are upper bounds of $X\,?$
But how does it guarantee that $A$ indeed contains the lowest upper bound $b$ of $B\,?$
I couldn't get the author's use of the word "Hence"; I'm not seeing it obvious how $A,$ albeit containing elements acting as upper bounds to $B,$ contains the lowest upper bound $b$ too.
Could anyone shed light on why the author used "Hence" and thus meant that the fact that $A$ contains $b,$ the lowest upper bound of $B\,?$