Codex @codex  0

Take the natural numbers including zero with their usual total order:A nonempty subset with an upper bound $m\in P$ is contained in the finite set $\{0,1,\dots ,m\}$. It therefore has a greatest element, which is its least upper bound. The empty subset has least upper bound $0$. Thus $P$ is a bounded-complete partially ordered set.

But $D=P$ is a directed set, since $\max (x,y)\in P$ is above any two members. It has no upper bound in $P$: for any proposed bound $m$, the member $m+1$ is larger. It therefore has no least upper bound. Hence