Why are singletons of Real Numbers an interval according to this definition? "Definition: A subset S of R is said to be an interval if it has the following property: if x∈S, z∈S, and y∈R are such that x < y < z, then y∈S"
According to the book, singletons are intervals from this definition, but since singletons only have one element, why would that be the case?
 A: Consider the singleton $S=[x,z]$ with $z=x$; i.e., $S=[x,x]=\{x\}$.
For this $S$, there is no $y\in\mathbb R$ such that $x<y<z=x$,
so the property in the definition of an interval is satisfied vacuously.
A: In order to prove that $S=\{p\}$ were not an interval, logically you have to give $x \in S$, $z \in S$ and some $y \in \Bbb R$ such that $x < y < z$ but nevertheless $y \notin S$:
(in first order logic the definition translates (quantifier domain is $\Bbb R$) as
$$\forall x,y,z: ((x \in S) \land (z \in S) \land (x < y) \land (y < z)) \to (y \in S)\tag{1}$$
which negates via standard rules (an implication is only falsified by having the antecedent true and the consequent false etc.)
$$\exists x,y,z: ((x \in S) \land (z \in S) \land (x < y) \land (y < z)) \land (y \notin S)\tag{2}$$
But as $x$ and $z$ are forced to be equal to $p$ but then $(p < y)$ and $y < p$ forces $p < p$ contradiction.
So $(2)$ (= not $(1)$) cannot hold for $S=\{p\}$ so because we have two-valued logic, $(1)$ does hold for $\{p\}$.
This is just an expansion of the vacuous truth notion that was already mentioned, in the hope it adds to the understanding of the why of it. You cannot falsify it so it's true.
