What is the name of a set $A$ that has its infimum $i \in A$ and a supremum $s \in A$?
Examples:
$[0,1]$ has a supremum $1$ and an infimum $0$ which are inside $[0,1]$.
$\{0,1,2,3\}$ has a supremum $3$ and an infimum $0$, both are inside.
$\{0,3\} \cup [1,2]$ has a supremum $3$ and an infimum $0$, both are inside.
Non-examples:
Any open interval does not have its infimum and a supremum inside.
$\mathbb{N}_0$ has an infimum $0$ which is inside it, but it does not have any supremum that would be inside it, i.e. the first infinite ordinal $\omega \notin \mathbb{N}_0$.
And what is the name of a poset whose all non-empty subsets have an infimum and a supremum inside?
Posets whose all non-empty subsets have an infimum and a supremum inside are exactly the finite totally ordered sets. The name of such a poset is "finite chain".
Indeed, let $(D, \leq)$ be a such poset. First, it is totally ordered: for any $x, y \in D$, consider the subset $\{ x, y \}$. Now, since all non-empty subsets have an infimum inside, $(D, \leq)$ is well ordered. In the same way, the dual of $(D, \leq)$ is well ordered. As an exercise, you can check that a well ordered set such that its dual is well ordered is finite.