De Morgan's laws are not necessary. We have
\begin{align}
x\in X\setminus\bigcup_{\alpha\in I} A_\alpha &\iff x\in X, x\notin A_\alpha \text{ for all $\alpha\in I$}\\
&\iff x\in X\setminus A_\alpha \text{ for all $\alpha\in I$}\\
&\iff x\in \bigcap_{\alpha \in I}(X\setminus A_\alpha),
\end{align}
and similarly for the other statement.
Also, De Morgan's laws hold for sets of arbitrary cardinality.
The property of being a good set depends on the function $s$ (the function that assigns every well-ordered set to a UNIQUE strict upper bound), the sets that you give in your example are not good sets, if you suppose that $Y$ and $Y'$ are good sets we get a contradiction: $8\in Y$ and $10\in Y'$ but since both sets are good sets then
$$8 = s(\{y\in Y : y < 8\}) = s(\{7\})$$
$$10 = s(\{y\in Y' : y < 10\}) = s(\{7\})$$
Here is the intuitive idea: The motivation behind a good set, is that are sets of the form:
$$X_1 = \{x_0\}$$
$$X_2 = \{x_0, s(X_1)\}$$
$$X_3 = \{x_0, s(X_1), s(X_2)\}$$
$$X_4 = \{x_0, s(X_1), s(X_2), s(X_3)\}$$
$$\vdots$$
So, if you take any two good sets
$$X_n=\{x_0, s(X_1),\dots, s(X_{n-1})\}$$
$$X_m=\{x_0, s(X_1),\dots, s(X_{m-1})\}$$
w.l.o.g suppose that $n < m$, then
$$X_m\backslash X_n = \{s(X_{n}),\dots, s(X_{m-1})\}$$
and clearly every $x\in X_m\backslash X_n$ is an strict upper bound for $X_n$.
Best Answer
This is how I would prove 8.5.13.
Let $\le$ be a transitive reflexive binary relation on $X,$ and $x_0\in X.$ As usual $x<x'$ means $x'\ne x\le x'.$ We will also assume that $\forall x,x'\in X\, (x\le x'\le x\implies x=x')$ (Such $\le$ are called strict) for reasons given in my final remark.
Let $W$ be the set of subsets of $X$ that are well-ordered by $<$ and have $x_0$ as their least member. We have $\{x_0\} \in W,$ so $W$ is not empty. We introduce a binary relation $<^*$ on $W$,
where $w<^*w'$ iff $$(i). w\subsetneqq w'$$ $$(ii). \forall x\in w \, \forall x'\in (w'\setminus w)\,(x<x').$$
We show there exists a $<^*$-maximal $Y\in W.$
From that, if $x'\in X$ is a $\le$-upper bound for $Y$ and $x'\not \in Y$ then by the strictness of $\le$ we have $\forall w\in Y\, (w<x').$ But then $Y'=Y\cup \{x'\}\in W$ with $Y<^*Y',$ contradicting the $<^*$-maximality of $Y$. So $Y$ has no strict upper bound.
Finding $Y$ is by the following, which those familiar with set theory will recognize as a common method:
A chain in $W$ is any $C\subset W$ such that $$(iii). \forall c,c'\in C\,(c<^*c'\lor c'<^*c\lor c=c').$$
Claim: If $C$ is a non-empty chain in $W$ then $\cup C\in W$ and $\cup C$ is a $<^*$-upper bound for $C.$
Proof of claim: If $x,x'\in \cup C$ then for some $c,c'\in C$ we have $x\in c$ and $x'\in c'.$ Now $C$ is a $<^*$-chain so $c''=c\cup c'$ is one of $c,c'$ by (i) and (iii), so $\{x,x'\}\subset c''\in C.$ And since $c''$ is well-ordered by $<$, we have $(x'<x\lor x<x'\lor x=x').$
$\bullet \;$ So (since, also, $\le$ is transitive), $\cup C$ is linearly ordered by $<.$
Now suppose $\emptyset \ne S\subset \cup C.$ Take $c_0\in C$ such that $c_0\cap S \ne \emptyset$ and let $$s_0=\min_<(c_0\cap S)$$ which exists because $c_0$ is well-ordered by $<.$
Now for any $s\in S,$ take $c\in C$ such that $s\in c.$
If $c<^*c_0$ or $c=c_0$ then $s\in c\subset c_0$ so $s\in c_0\cap S,$ so by definition of $s_0$ we have $(s_0<s\lor s_0=s).$
Or if $c_0<^*c$ then
$(a).$ If $s\not \in c_0$ then by $(ii)$ we have $\forall w\in c_0 \,(w<s ),$ and in particular $s_0\in c_0,$ so $s_0<s.$
$(b).$ If $s\in c_0$ then $s\in c_0\cap S$ so by definition of $s_0$ we have $(s_0<s \lor s_0=s).$ (Similar to a previous case above.)
$\bullet \bullet \;$ So $\cup C$ is well-ordered by $<,$ so $\cup C \in W.$
Finally if $c\in C$ and $s\in (\cup C) \setminus c,$ take $c'\in C$ with $s\in c'.$ We cannot have $c'=c$ but we cannot have $c'<^* c$ either, else $s\in c'\subset c.$ So $c<^*c'$ and $s\in c'\setminus c,$ so by $(ii)$, we have $\forall w\in c\, (w<s).$ So $c<^*\cup C\lor c=\cup C.$
$\bullet \bullet \bullet \;$ So $\cup C$ is a $<^*$-upper bound for $C.$ QED.
Now by the claim (and since $W$ is not empty) we can apply Zorn's Lemma to $(W,<^*)$ and conclude there exists a $<^*$-maximal $Y\in W.$
Remark. $Y$ need not be $\subset$-maximal nor unique For example if $X=\{x_0,x_2,x_3\}$ with $x_0<x_1<x_2$ then $X$ and $\{x_0,x_3\}$ are $<^*$-incomparable and $<^*$-maximal
Final remark. If the poset is not assumed to be strict we might have $X=\{x_0,x_1\}$ with $x_0\ne x_1$ and $x_0\le x_1\le x_0.$ Then $Y=\{x_0\}$ but $x_1$ is a $\le$-strict upper bound for $Y$.