A Scott-like continuity between partially-ordered sets Given two partially ordered sets $P$ and $Q$, a function $f : P → Q$ between them is Scott-continuous if it preserves all directed suprema, i.e. if for every directed subset $D$ of $P$ with supremum in $P$ its image has a supremum in $Q$, and that supremum is the image of the supremum of $D$: that is, $⊔f[D] = f(⊔D)$, where $⊔$ is the directed join.
Is there a name for a the following variation on Scott continuity that does not make use of directed sets?

Given two partially ordered sets $P$ and $Q$, a function $f: P → Q$ between them is continuous if it preserves all suprema, i.e. if for every subset $D$ of $P$ with supremum in $P$ its image has a supremum in $Q$, and that supremum is the image of the supremum of $D$: that is, $\sup f[D] = f(\sup D)$.

 A: Your question "Is there a name for a the following variation on Scott continuity [...]" is somewhat loaded, because the property of functions between posets that you define cannot be called continuity, because, unless you restrict the class of posets to which it is supposed to be applicable, it cannot be equivalent to any continuity.
You propose to call a function "continuous" in some sense if and only if it preserves all suprema.  In particular, such "continuous" functions will have to preserve the supremum of the empty set, that is, the least element (the bottom) of the poset.  However, this clearly contradicts the fact that every constant function is continuous, whatever the topologies are.  Thus, you definition cannot be equivalent to continuity.
(However, if, for example, you restrict the definition to the class of singleton posets -- posets consisting of a single element each -- then, of course, the only function between each two such posets will be continuous if and only if it will satisfy this "variation on Scott continuity.")
This was easy to answer by considering the suprema of the empty set, so let us look at a modified version of your proposed definition of some "continuity":

Call a function between posets "continuous" if and only if it preserves the suprema of all non-empty subsets.

This definition is also absurd.  To see it, consider the two posets:
$$
  P =\{a,b,c\}\ \text{with $a > b$, $a > c$},\qquad
  Q =\{d, e\}\ \text{with d > e}.
$$
P:    a                 Q:  d
     / \                    |
    b   c                   e

Since the function $Q\to Q$ that exchanges its two elements does not "preserve" the supremum of $Q$, it is not "continuous."  This excludes the discrete and the anti-discrete topologies on $Q$.  The only two possible topologies on $Q$ left are:
$$
  \mathcal{T}_d =\{\varnothing,\{d\},Q\}
  \quad\text{or}\quad
  \mathcal{T}_e =\{\varnothing,\{e\},Q\}.
$$
The function
$$
  f\colon P\to Q,\quad
  a\mapsto d,\quad
  b, c\mapsto e
$$
is not "continuous," since it does not "preserve" the supremum of $\{b, c\}$, while the functions
$$
  g\colon P\to Q,\quad
  a, b\mapsto d,\quad
  c\mapsto e
$$
and
$$
  h\colon P\to Q,\quad
  a, c\mapsto d,\quad
  b\mapsto e
$$
are "continuous."
Suppose the topology on $Q$ is $\mathcal{T}_d$ (this is the Scott topology, by the way).  Then, since $g$ and $h$ are continuous, $\{a, b\}$ and $\{a, c\}$ are open in $P$, hence so is $\{a\}$, and therefore $f$ is continuous -- contradiction.  Thus the topology on $Q$ is not $\mathcal{T}_d$.
The only possible topology on $Q$ left is $\mathcal{T}_e$.  Since $g$ and $h$ are continuous, $\{b\}$ and $\{c\}$ are open in $P$, hence so is $\{b, c\}$, and therefore $f$ is continuous -- contradiction.  Thus, the proposed definition of "continuous" functions is absurd.
A: Why not say that $f$ preserves existing supremum and, if you feel it is necessary, provide the defining formula.
