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Use the two-chain noncontinuous complete partial orderwhere $\perp <a_1<a_2<\cdots <\top$ and $\perp <b_1<b_2<\cdots <\top$, with no comparisons between the two interiors. In this solution $\sup D$ denotes the least upper bound printed as $\sqcap D$ in the question.^[ref]

Every nonempty directed set containing $\top$ has least upper bound $\top$. A directed set omitting $\top$ cannot contain both an $a_i$ and a $b_j$, because their only common upper bound in $P$ is $\top$, which would have to belong to that directed set. It therefore lies in one chain, possibly with $\perp$. If its indices are bounded it has a greatest element, which is its least upper bound; if they are unbounded its least upper bound is $\top$. Thus $P$ is a directed-complete partial order.

No element other than $\perp$ is way-below $\top$. For $a_i$, the directed set $\{b_n:n\ge 1\}$ has least upper bound $\top$, but none of its members is above $a_i$. Interchanging the two chains excludes every $b_i$. Either chain excludes $\top$ itself. On the other hand, $\perp \ll \top$: every directed set whose least upper bound is above $\top$ is nonempty, and any member is above $\perp$. HenceSo $P$ is not a continuous directed-complete partial order.

There is a small convention issue in the printed definitions. Our usual directed set is nonempty, but the question's pairwise condition also allows the empty set. Taken literally, the empty set has least upper bound $\perp$, and the preceding example and failure of continuity still work. Under this literal convention $\perp /\ll \perp$, because the empty set supplies no witness. We retain this qualification in the next parts.