The statement here is to be proved beyond a shadow of a doubt, but in lectures, we made an intuitive conclusion.
I think infimum is the set of the least cardinality which contains common elements of $N,O,P,R \in S,$
while supremum is the set of the greatest cardinality which contains all the elements of each of $N,O,P$ and $R$
$(\mathcal P(S), \leq)$ $$S=\{\underbrace{\{a,b,c,d\}}_N,\underbrace{\{b,c,d,e\}}_O,\underbrace{\{a,c,d,e\}}_P,\underbrace{\{c,d,e,f\}}_R\}$$ $\inf A=N\cap O\cap P\cap R\ =\ \{c,d\}$
$\sup A=N\cup O\cup P\cup R\ =\ \{a,b,c,d,e,f\}$
In my view, cardinality is the spectrum, like an interval on the number line. Please, correct me if I am wrong.
First, you've got it backwards. The set $\inf A$ is the subset of $S$ of greatest cardinality that contains common elements of $N,O,P,R$. For example, the set $\{c\}$ contains common elements of $N,O,P,R$ but it has less cardinality than $\{c,d\}$. Similarly, $\sup A$ is the set of least cardinality which contains all the elements of each of $N,O,P,R$. This is not unlike the concepts of infimum and supremum in the number line: given a bounded subset $X \subset \mathbb R$, $\inf X$ is the greatest lower bound of $X$, and $\sup X$ is the least upper bound of $X$.
Second, your cardinality statement only works when $S$ is a finite set. All bets are off if $S$ is infinite. For example, suppose $S = \mathbb N$, the natural numbers, and suppose that $N$ is all multiples of $2$, $O$ is all multiples of $3$, $P$ is all multiples of $5$, and $R$ is all multiples of $y$. Then all of the sets $N$, $O$, $P$, $R$, $N \cap O \cap P \cap R$, and $N \cup O \cup P \cup R$ have the same cardinality, which is equal to the cardinality of $\mathbb N$ itself. There are subsets of $\mathbb N$ that are properly contained in $N \cap O \cap P \cap R$ that have the same cardinality, and there subsets properly containing $N \cup O \cup P \cup R$ that have the same cardinality. To be precise at least for the intersection, $N \cap O \cap P \cap R$ is all multiples of $2 \cdot 3 \cdot 5 \cdot 7 = 210$, and it has "smaller" subsets of the same cardinality: just omit the first element $210$, leaving in $2 \cdot 210 = 420$ and every higher multiple of $210$.