Why does this implication about this supremum in Axler’s proof of the Heine-Borel theorem hold?

I have a question about the proof of the Heine-Borel theorem in Sheldon Axler’s Measure, Integration & Real Analysis. The theorem and the associated proof are reproduced below.

I don’t understand why $s=\sup D$ implies the existence of $d \in (s - \delta, s]$ and $n \in \mathbb{Z}^+$ and $G_1, \ldots, G_n \in \mathcal{C}$ such that$$ [a,d] \subset G_1 \cup \cdots \cup G_n.$$

I’ve been thinking about this for some time now but can’t seem to figure it out. Thanks for considering my question!

Heine-Borel Theorem. Every open cover of a closed bounded subset of $\mathbb{R}$ has a finite subcover.

Proof. Suppose $F$ is a closed bounded subset of $\mathbb{R}$ and $\mathcal{C}$ is an open cover of $F$.

First consider the case where $F = [a,b]$ for some $a,b \in \mathbb{R}$ with $a < b$. Thus $\mathcal{C}$ is an open cover of$[a,b]$. Let $$ D=\{d\in [a,b] : [a,d] \text{ has a finite subcover from } \mathcal{C}\}. $$ Note that $a \in D$ (because $a \in G$ forsome $G \in \mathcal{C}$). Thus $D$ is not the empty set. Let $$ s = \sup D. $$ Thus $s \in [a,b]$. Hence there exists an open set $G \in \mathcal{C}$ such that $s \in G$. Let $\delta > 0$ be such that $(s -\delta, s + \delta) \subset G$. Because $s = \sup D$, there exist $d \in (s - \delta, s]$ and $n \in \mathbb{Z}^+$ and $G_1, \ldots, G_n \in \mathcal{C}$ such that $$ [a,d] \subset G_1 \cup \cdots \cup G_n. $$ Now $$ [a,d'] \subset G \cup G_1 \cup \cdots \cup G_n $$ for all$d' \in [s, s + \delta)$. Thus $d' \in D$ for all $d' \in [s, s + \delta) \cap [a,b]$. This implies that $s = b$. Furthermore, the argument with $d' = b$ shows that $[a,b]$ has a finite subcover from $\mathcal{C}$, completing the proof in the case where $F = [a,b]$.

Now suppose $F$ is an arbitrary closed bounded subset of $\mathbb{R}$ and that $\mathcal{C}$ is an open cover of $F$. Let $a, b \in \mathbb{R}$ be such that $F \subset [a,b]$. Now $\mathcal{C} \cup \{\mathbb{R} \setminus F\}$ is an open cover of $\mathbb{R}$ and hence is an open cover of $[a,b]$ (here $\mathbb{R} \setminus F$ denotes the set complement of $F$ in $\mathbb{R}$). By our first case, there exist $G_1, \ldots, G_n \in \mathcal{C}$ such that $$ [a,b] \subset G_1 \cup \cdots \cup G_n \cup (\mathbb{R} \setminus F). $$ Thus $$ F \subset G_1 \cup \cdots \cup G_n, $$ completing the proof.