Let $P$ be a partially ordered set, and let $I(P)$ be the set of order ideals (i.e. downward-closed sets) in $P$ ordered by inclusion. Suppose that $X$ is a chain of non-empty elements of $I(P)$. Claim: there is a monotonic function $f : X \to P$ such that $f(J) \in J$ for each $J \in X$. In other words, the chain of ideals has a corresponding chain of elements. Is this claim true?
Edit. To clarify, by "monotonic" I don't mean "strictly monotonic". That is, we just want for $J, K \in X$, if $J \subseteq K$ then $f(J) \leq f(K)$.
Counterexample: Let $$P=\{(x,y)\in\omega\times\omega:x\ge y\}$$ with the ordering $$(x',y')\le(x,y)\iff x'=x\wedge y'\le y$$ and let $X=\{J_n:n\in\omega\}$ where $$J_n=\{(x,y)\in P:x\ge y+n\}.$$
Simpler example: Let $P=\omega$ with the trivial ordering $$x\le y\iff x=y$$ and let $X=\{J_n:n\in\omega\}$ where $$J_n=\{n,\,n+1,\,n+2,\,n+3,\dots\}.$$