As mentioned in Uncountable chains (in the question itself), one can do this as follows: Pick a bijection between $\mathbb{N}$ and $\mathbb{Q}$ so it suffices to find such a chain in the powerset of $\mathbb{Q}$. But $\mathbb{Q}\subseteq\mathbb{R}$ so for any real number $r$ we can define the su...

$P(\mathbb N)$ = power set of $\mathbb N$. $A \subset P(\mathbb N)$ is a chain if $a,b \in A \implies$ either $a \subseteq b$ or $ b \subseteq a$ That is we have something like this: $$\ldots a \subseteq b \subseteq c \subseteq\ldots$$ where $a,b,c \in A$ are distinct. We can show easy enough...

Let $B$ be some set. The problem is to find a set $A\subset\mathcal{P}(B)$ of subsets of $B$ which is totally ordered by inclusion and such that there exists a bijection $A\leftrightarrow \mathcal{P}(B)$. This is an easy exercise if $B$ is countable where one can explicitly construct such a set ...