Existence of an increasing subsequence of sets that preserves supremum Background
This is not necessary to understand the question
The following corresponds to a claim made on page 7 of the book Optimal Stopping and Free-Boundary Problems, Peskir and Shiryaev 2006.
Let $I$ be any (index) set and let $\lbrace Z_\alpha |\, \alpha \in I \rbrace$ be a family of random variables on $(\Omega, \mathscr{G}, \mathbb{P})$. Let $\mathcal{C}$ denote the set of all countable subsets of $I$. It is claimed that there exists an increasing sequence $\lbrace C_n \, | \, n \geq 1 \rbrace$ in $\mathcal{C}$ such that
$$
\sup_{C \in \mathcal{C}} \mathbb{E}[\sup_{\alpha \in C} Z_\alpha] = \sup_{ n \ge 1 } \mathbb{E}[\sup_{\alpha \in C_n} Z_\alpha]
$$
This claim is implicitly made in a proof that I would like to understand. This is the proof of a Lemma called Essential Supremum. I think the claim follows from the following Lemma that I have formulated myself.
Lemma
Let $I$ be any set and let $\mathcal{C} \subset \mathscr{P}(I)$ and let $f: \mathcal{C} \rightarrow \mathbb{R}$ be increasing in the sense that $A \subseteq B \implies f(A) \leq f(B)$. Then there exists an increasing sequence $D = (D_n)_{n=0}^\infty$ with $D_n \in \mathcal{C}$ such that
$$
\sup_{C \in \mathcal{C}} f(C) = \sup_{ C \in D} f(C)
$$
Proof
Let $a = \sup_{C \in \mathcal{C}} f(C)$. 
Assume that $a < \infty$. Let $b_n = a - 1/n$. We construct a sequence $B = (b_n)_{n=0}^\infty$ as follows. For every $n \in \mathbb{N}$, $b_n < a$, so $b_n$ is not an upper bound on $f(\mathcal{C}) = \lbrace f(C) \,|\,\, C\in \mathcal{C} \rbrace$, so we can choose $B_n$ in $\mathcal{C}$ such that $f(B_n) \geq b_n$. Now we construct $D$ by setting, for all $n \in \mathbb{N}$, $D_n = \bigcup_{k=1}^n B_k$ so that because $f$ is increasing we have $f(B_n) \le f(D_n)$. For any $c < a$ we have that there exists an $n \in \mathbb{N}$ such that $c < b_n < f(B_n) \le f(D_n) $, so $c$ is not an upper bound on $f(D) = \lbrace f(D_n) \,|\,\, n \in \mathbb{N} \rbrace$. Because $D \subset \mathcal{C}$, and $a$ is an upper bound on $f(\mathcal{C})$, $a$ is an upper bound on $f(D)$. So $a \le \sup \lbrace f(D_n) \, | \,\, n \in \mathbb{N} \rbrace \le a$, so $\sup \lbrace f(D_n) \, | \,\, n \in \mathbb{N} \rbrace = a$. 
Now assume that $a = \infty$. We set $b_n = n$ and make the same construction. There is no upper bound on $b_n$, so there is no upper bound on $f(D)$, so $\sup \lbrace D_n \, | \,\, n \in \mathbb{N} \rbrace = \infty =a$. 
In both cases $D$ is increasing by construction, as $D_{n+1} = D_n \cup B_{n+1}$.
Questions
I originally thought that in the Lemma $\mathcal{C}$ would need to have countable elements. Is the Lemma true without this assumption, as I have formulated it (i.e. is the proof correct)?
Are there any useful generalisations of my Lemma? For example is it still true if we replace the system of sets $\mathcal{C}$ with its partial order $\subset$, as well as $\mathbb{R}$ with its total order $\leq$ by general objects? (Perhaps I should not care too much about this yet)
Does my Lemma have a name? I thought I might find it in the book Mathematical Analysis I, Canuto and Tabacco 2015, but I could not find it directly.
 A: Answer to first question
The proof is not correct, the requirement that $\mathcal{C}$ is the collection of countable subsets of $I$, or more generally that it is closed under countable unions, is needed in this proof. The mistake in the proof is that while $B_n$ is an element of $\mathcal{C}$, $D_n$ need not be an element of $\mathcal{C}$. 
A counterexample is this. Let $I = \mathbb{Z}$ let $C_n = \lbrace -n \rbrace \cup \lbrace 0, \dots, n \rbrace$ and with usual abuse of notation $\mathcal{C} = (C_n)_{n=1}^\infty = \lbrace C_n \, | \,\,n \ge 1 \rbrace$. Furthermore, let $f:\mathcal{C} \rightarrow \mathbb{R} : C \mapsto \max(C)$. 
We have 
$$
a= \sup_{C\in\mathcal{C}} f(C) = \sup_{n\geq1} f(C_n) = \sup_{n\geq1} n = \infty
$$
The increasing sequences of $\mathcal{C}$ have only one element, because by construction no element of $\mathcal{C}$ is a proper subset of another. So for any increasing sequence $D$ we have $D = \lbrace C_n \rbrace$ for some $n$ and
$\sup_{C \in D} f(C) = f(C_n) = n$
So there does not exist an increasing sequence $D$ for which the suprema are equal.
