Here is the excerpt from the textbook A Course in Mathematical Analysis by Prof D. J. H. Garling.
Here, $P(A)$ is the power set of $A$ and "increasing" means relation to inclusion. i.e. $X\subseteq Y\implies f(X)\subseteq f(Y)$. $s(n)$ is the successor function. i.e. $s(n)=n+1$.
From $H=\cup_{n=0}^{\infty}H_{n}=H_{0}\cup (\cup_{n=1}^{\infty}H_{n})=\varnothing\cup(\cup_{n=1}^{\infty}H_{n})=\cup_{n=1}^{\infty}H_{n}$.
In short, $H=\cup_{n=1}^{\infty}H_{n}$.
$S(H)=S(\cup_{n=0}^{\infty}H_{n})=\cup_{n=0}^{\infty}S(H_{n})=S(H_{0})\cup S(H_{1})\cup S(H_{2})\cup S(H_{3})...=H_{1} \cup H_{2} \cup H_{3} \cup H_{4} ... =\cup_{n=1}^{\infty}H_{n}$.
In short, $S(H)=\cup_{n=1}^{\infty}H_{n}$.
To sum up, we have $H=\cup_{n=1}^{\infty}H_{n}$ and $S(H)=\cup_{n=1}^{\infty}H_{n}$. This implies $H = S(H)$.
If my reasoning is correct. Question (d) will be wrong.
Please have a check!
The problem with your reasoning is that you confuse evaluation with image.
Recall that if $F\colon X\to Y$ is a function, we often write $F(A)=\{F(a)\mid a\in A\}$ as a subset of $Y$. But if your function is from sets to sets, then $F(A)$ is an element of $Y$, and not [necessarily] a subset of $Y$.
This is one of the reasons that in set theory you would see $F[A]$ or $F"A$ used to denote $\{F(a)\mid a\in A\}$.
In turn this implies that you are mistaken about the continuity of $S$ here. There is no reason for $S(\bigcup H_n)$ to be equal to $\bigcup S(H_n)$. As bof suggests in the comments, take $A$ to be a well-ordered set which is large enough (e.g. $\{\frac{n}{n+1}\mid n\in\Bbb N\}\cup\{1\}$) and $S(H)=H\cup\{a\}$ for $a$ as the least element not in $H$; and $S(A)=A$.
Then $H_0=\varnothing$, and $H_{n+1}=S(H_n)$ will give you finite initial segments of the order. So $\bigcup H_n$ will be just $\{\frac n{n+1}\mid n\in\Bbb N\}\neq A$, but $S(\bigcup H_n)=A$ itself.