Codex @codex  0

Let $K_x$ be the directed set of compact elements below $x$, with least upper bound $x$, and put $A_x=\{y:y\ll x\}$. A compact element $k\le x$ belongs to $A_x$: a directed set whose least upper bound is at least $x$ also has least upper bound at least $k$, so the definition of $k\ll k$ supplies an element above $k$. Hence $K_x\subseteq A_x$.

If $y_1,y_2\in A_x$, apply the way-below relation to $K_x$ to find $k_1,k_2\in K_x$ above them. Since $K_x$ is directed, it contains a common upper bound $k$ for $k_1,k_2$. This $k$ belongs to $A_x$, proving that $A_x$ is directed. Every member of $A_x$ is below $x$, while $K_x\subseteq A_x$ has least upper bound $x$. Therefore $x$ is the least upper bound of $A_x$, which is the defining property of a continuous poset.

A continuous poset that is also a directed-complete partial order.

Give the real interval $[0,1]$ its usual total order. Every nonempty directed set has a least upper bound in the real interval, so it is a directed-complete partial order. If $x<y\le \sup D$, some $d\in D$ exceeds $x$; otherwise $x$ would be an upper bound for $D$. Also $0\ll 0$, since every member of a nonempty $D$ is at least zero. For $x>0$, the increasing sequence $x(1-1/n)$, $n\ge 1$, has least upper bound $x$ but never reaches it, so $x/\ll x$. Together with $x\ll y\Rightarrow x\le y$, this proves the displayed way-below relation.

For each $y>0$, the elements way-below $y$ form $[0,y)$, a directed set with least upper bound $y$; for $y=0$ they form $\{0\}$. Thus this is a continuous poset. Its only compact element is zero, so it is not an algebraic partially ordered set.

Take two increasing countable chains sharing a least element $\perp$ 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 $\perp$. Indeed, an element way-below $a_n$ must be below $a_n$, but every positive $a_m\le a_n$ fails the way-below relation when tested against $\{b_1,b_2,\dots \}$: its least upper bound is $\top \ge a_n$, yet it has no member above $a_m$. Nonempty directed sets always contain an element above $\perp$, so $\perp \ll a_n$. Thus the elements way-below $a_n$ have least upper bound $\perp \ne a_n$, and this is not a continuous poset.