How can I show that given a partially ordered set $(A\le)$ with bottom, the set $\operatorname{dc}(A)$ that contains all subsets of $A$ that are down-closed is chain-complete?
By "partially ordered set" I mean a set with a partial order relation $\le$ over which is reflexive, antisymmetric, and transitive, i.e., which satisfies for all $a, b,$ and $c$ in $P$: $a\le a$ (reflexivity: every element is related to itself); if $a\le b$ and $\le a$, then $a = b$ (antisymmetry: there exists at most one relation between two distinct elements); if $\le b$ and $b\le c$, then $a\le c$ (transitivity: if a first element is related to a second element, and, in turn, that element is related to a third element, then the first element is related to the third element).
By "bottom" I mean the least element of the set.
By "down-closed" I mean that for each element $x$ in a set $P\subseteq A$ if $y\in A$ and $y\le x$ then $y\in P$.
By "chain-complete" I mean a poset in which each chain has a least upper bound.
By "chain" I mean a set in which for all elements $x,y\in the chain we have either $x\le y$ or $y\le x.$
$\operatorname{dc}(P)$ is ordered by set inclusion.
Consider a chain $\mathcal C$ in $\operatorname{dc}(A);$ we have to show that $\mathcal C$ has a least upper bound in $\operatorname{dc}(A).$
Let $U=\bigcup\mathcal C,$ the union of all the members of $\mathcal C.$
I claim that $U\in\operatorname{dc}(A),$ i.e., that $U$ is down-closed. Suppose $y\in A$ and $y\le x\in U;$ I have to show that $y\in U.$ Since $x\in U=\bigcup U,$ we have $x\in P$ for some set $P\in U.$ Since $P$ is down-closed, we have $y\in P\subseteq U,$ so $y\in U.$
Clearly, $U=\bigcup\mathcal C$ is the least upper bound of $\mathcal C$ in $\operatorname{dc}(A).$