The correct title must be: Proof of Theorem 90 of Appendix of Kelley's book General Topology or, Proving a theorem about $\mathcal R$-sections or Prove of this prove that proof aboutTheorem 90 of Appendix of Kelley's book General Topology .
Proposition. Let $(X,\mathcal R)$ be an ordered set (or class). If $Y\subset X$ is an $\mathcal R$-section and every element $y\in Y$ is also an $\mathcal R$-section of $X$, then $\bigcup Y$ and $\bigcap Y$ are $\mathcal R$-sections of $X$.
EDIT 1: The statement was misunderstood by me. The real porposition is:
Let $I$ non-empty (possibly $I$ is a set). If every element $y\in I$ is an $\mathcal R$-section of $X$, then $\bigcup I$ and $\bigcap I$ are also.
So the comment of @WilliamElliot is unnecesary, because in any momment we have said $I\subset X$. And probably it may be false. In my opinion, $I$ plays the role of an index set.
Addendum 1. This statement corresponds to the Appendix of J. L. Kelley General Topology, Theorem 90 (p. 264).
EDIT 2: Below is the rest of my question. Whit the new remarks it has no sense. My problem is that I can't understand why (or how) an element $y\in Y$ can be an $\mathcal R$-section of $X$ if $y$ is not a subset of $X$ (as least as far as I know). It would have sense for me if the text said $\{y\}$, which is actually a subset of $X$.
The same applies to $\bigcup Y$ and $\bigcap Y$: by DEFINITION, both are sets (classes) formed by elements of sets in $Y$:
$$ \bigcup Y= \{x:\exists y\in Y \mbox{ such that } x\in y\}\\ \bigcap Y= \{x|x\in y \;\forall y\in Y\}. $$
Again, to be an $\mathcal R$-section, we should consider $\{\bigcup Y\}$ and $\{\bigcap Y\}$.
Addendum 2. (Real questions) Once I have understood the proposition, the result seems me evident. However, I don't know how how to prove it rigorously. I know I have a collection of sets $\{y:y\in I\}$ such that
$$ \mbox{if } y_1,y_2\in Y \mbox{ with } y_1\neq y_2 \Longrightarrow y_1\subset y_2 \mbox{ or } y_2\subset y_1 , $$
So I actually have a descendent chain of sets
$$ y\supset y' \supset y'' \supset \cdots $$
1.- Proof for $\bigcup I$. With the above in mind, I think $bigcup I = y$ which is an $\mathcal R$-section by hypothesis.
2.- Proof for $\bigcap I$. I think the definition of $\mathcal R$-section avoids there is some $y\in I$ such that $y=\emptyset$. So I think all them has a common element, the least element $y_0$, and thus $\bigcap I$ is non-empty. Now, let $x\in X$ and $y\in\bigcap I$ with $x\mathcal R y$ and suppose $x\notin \bigcap I$. That implies that $x$ is less than the least element $y_0$. But then, $x$ wouldn't below either to $y\in I$, which is a contradiction, because $y$ was an $\mathcal R$-section$. So $\bigcap I$ is an $\mathcal R$-section.
Questions about this proof:
1.- Is the proof of $bigcup I$ correct?
2.- Is the proof of $\bigcap I$ correct?
3.- In 2., do you think I need to proove $y_0\in \bigcap I$? Is there a shorter proof?
4.- If all the inclusions are proper, then is it possible to show that $\bigcap I=\{y_0\}$?
Definition. Let $\mathcal R$ be a well-order in $X$. An $\mathcal R$-section is $Y\subset X$ such that if $x\in X$ and $y\in Y$ with $x\mathcal R y$, then $x\in Y$.
Informally, a set (class) $Y$ is said to be an $\mathcal R$-section if there is no element in $X\setminus Y$ that precedes the elements of $Y$.
Final Remark. I have prefered keep the old question and add the edits and corrections, because I think that is more proper. Sorry if it is a bad idea.
Thanks a lot for your patience.
Well I think the notations in the book have been misunderstood. I have the 1964 edition, in which the theorem is,
$90$ THEOREM If $n\not=0$ and each member of $n$ is an $r$-section of $x$ , then $\bigcup n$ and $\bigcap n$ are $r$-sections of $x$.
Here $0$ is the void class, and $n$ is a class-of-classes; this may be understood from the definition of the union $\bigcup$ and the intersection $\bigcap$ operators.
Let $n=\{\{1,2\},\{1,2,5\},\{1,2,3,4,5\} \}$. Then, $\bigcup n=\{1,2,3,4,5\}$ and $\bigcap n = \{1,2\}$.
In this sense, the theorem can be reworded in terms of ordinary set theory as:
If $Y_i$, where $i=1,2,3,...$ be $r$-sections of a set $X$, so are $\cup_iY_i$ and $\cap_iY_i$.
A subset $Y\subset X$ is an $r$-section of $X$ iff the relation $r$ well orders $X$ and there is no $u\in (X-Y)$ such that $u\ r\ v$ for any $v\in Y$.
Using the definition alone the theorem appears logically obvious. A formal proof I leave on the author.