Example 13, Sec. 3 in Munkres' TOPOLOGY, 2nd ed: Necessary and Sufficient Conditions for a Subset to Have the Least Upper Bound Property Let $A$ be an ordered set having the least upper bound property, and let $A_0$ be a non-empty subset of $A$. 
Then what is (are) the necessary and sufficient condition(s), if any, for $A_0$ to have the least upper bound property under the induced ordered relation? 
In the example in Munkres, the subset $(-1, 1)$, for example, of $\mathbb{R}$ does have the least upper bound property, whereas the subset $(-1, 0)\  \cup \ (0, 1)$ does not. 
Now we can say with certainty that every (open, closed, bounded, or unbounded) interval in $\mathbb{R}$ does have the least upper bound property. Am I right?
Can we assert anything beyond that for $\mathbb{R}$ for a general ordered set having the least upper bound property?

Incomplete answer.
Let $B\subseteq A_{0}$ be non-empty (with $y\in B$) and bounded
above in $A_{0}$ (having $x\in A_{0}$ as upperbound). 
Let $z$ denote the least upper bound of $B$
looking at it as a subset of the original $A$. 
Then $y\leq z\leq x$. 
So if $A_{0}$
has the property $$y,z\in A_{0}\implies\left[y,x\right]\subseteq A_{0}\tag1$$
then the least upperbound of $B$ in $A$ will be an element of $A_{0}$. 
Evidently the least upper bound in the original order will also serve as least upperbound in $A_0$.
Apparantly (1) is a sufficient condition for $A_{0}$ to have the
least upper bound property. Intervals have this property. I am afraid that the property is even characteristic for intervals. However, there are subsets of $\mathbb R$ that are not an interval and do have the least upper bound property. Example: $(0,1)\cup\{2\}$. So there is no necessity.
