= Two-chain noncontinuous complete partial order
Take two increasing countable <chains> sharing a <least element> $\bot$ and a <greatest element> $\top$, with interiors $a_1<a_2<\cdots$ and $b_1<b_2<\cdots$ and no comparisons between branches. Each entire interior <chain> has $\top$ as its <least upper bound>. A <directed set> meeting both interiors must contain $\top$. Otherwise it lies within one branch, where it has either a greatest member or <least upper bound> $\top$. This proves that the <partially ordered set> is a <directed-complete partial order>.
The only element <way-below> $a_n$ is $\bot$. Indeed, an element <way-below> $a_n$ must be below $a_n$, but every positive $a_m\leq a_n$ fails the <way-below relation> when tested against $\{b_1,b_2,\ldots\}$: its <least upper bound> is $\top\geq a_n$, yet it has no member above $a_m$. Nonempty <directed sets> always contain an element above $\bot$, so $\bot\ll a_n$. Thus the elements <way-below> $a_n$ have <least upper bound> $\bot\ne a_n$, and this is not a <continuous poset>.
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