In an arbitrary ordered field one has the notion of Dedekind completeness -- every nonempty bounded above subset has a least upper bound -- and also the notion of sequential completeness -- every Cauchy sequence converges.
The main theorem here is as follows:
For an ordered field $F$, the following are equivalent:
(i) $F$ is Dedekind complete.
(ii) $F$ is sequentially complete and Archimedean.
Since the Archimedean hypothesis often goes almost without saying in calculus / analysis courses, many otherwise learned people are unaware that there are non-Archimedean sequentially complete fields. In fact there are rather a lot of them, and they can differ quite a lot in their behavior: e.g. some of them are first countable in the induced (order) topology, and some of them are not. (In particular there are some ordered fields in which a sequence is Cauchy if and only if it is eventually constant! This is a case where one should consider Cauchy nets if one is serious about exploring the topology...)
These issues are treated in $\S 1.7$ and $1.8$ of these notes.
We assume that we know the Least Upper Bound Principle: Every non-empty set $A$ which is bounded above has a least upper bound.
We want to show that every non-empty set $B$ which is bounded below has a greatest lower bound.
Let $A=-B$. So $A$ is the set of all numbers $-b$, where $b$ ranges over $B$.
First we show that if $B$ has a lower bound, then $A$ has an upper bound. For let $w$ be a lower bound for $B$. We show that $-w$ is an upper bound for $A$.
Because $w$ is a lower bound for $B$, we have $w\le b$ for any $b\in B$. Multiply through by $-1$. This reverses the inequality, and we conclude that $-w\ge -b$ for any $b\in B$. The numbers $-b$ are precisely the elements of $A$, so $-w\ge a$ for any $a\in A$.
Since $A$ is bounded above, it has a least upper bound $m$. We show that $-m$ is a greatest lower bound of $B$.
As usual, the proof consists of two parts (i) $-m$ is a lower bound for $B$ and (ii) nothing bigger than $-m$ is a lower bound for $B$.
The proofs again use the fact that multiplying by $-1$ reverses inequalities. To prove (i), suppose that $-m$ is not a lower bound for $B$. Then there is a $b\in B$ such that $b\lt -m$. But then $-b\gt m$. Since $-b\in A$, this contradicts the fact that $m$ is an upper bound of $A$.
To prove (ii), one uses the same strategy.
Best Answer
The answers to your two questions are "yes" and "yes".
First, this statement is one of the standard axioms for real numbers, called the "completeness axiom".
Second, in the standard "Dedekind cut" construction of the real numbers, one starts from an axiomatic description of the rational numbers, and then one constructs the real numbers and proves that all of their axioms hold, including the completeness axiom.