I know that this question is pretty old, but since it has reappeared I take the opportunity to give a justification of this axiom due to Dana Scott (I think).

First, some general remarks. I understand that to be justified is to be justified from a list of principles characterizing the notion of set-formation set theory is supposed to be about. The old cantorian notion of set-formation as unlimited/undisciplined/unrestricted gathering leads to paradoxes and has long been replaced by an ordered/organized/disciplined set-formation which became the standard conceptual basis of set theory. This new notion can be characterized by the following principles:

1) Sets ought to be determined by their elements and produced in ordered stages without upper bound. If a set is produced at some stage, then each of its elements ought to be produced at an earlier stage.

2) At each stage any plurality of sets produced in previous stages ought to determine a set.

3) Given a set and stages functionally connected to its elements, there ought to be a stage after all those given stages.

There is no need to demand that the stages are well-ordered, contrary to a common belief.
Now, Dana Scott's proof:

First, a set $x$ is said to be grounded if and only if every set containing $x$ as an element has a minimal element. The relation between sets and stages, that $x$ is produced at stage $s$, is denoted by $x R s$, and the ordering of stages is denoted by $\triangleleft$.

If all elements of a set $x$ are grounded, then $x$ is also grounded. For, suppose $x$ is an element of $y$. Then, either $y$ and $x$ have no common member, in which case $x$ is minimal, or there is a $z$ such that $z \in y$ and $z \in x$. From $z \in x$, it follows that $z$ is grounded, and from $z \in y$, it follows that $y$ has a minimal element.

If $s$ is a stage, then, from the second principle, there is a unique set $G_s$ which is produced at stage $s$ and whose elements are exactly the grounded individuals related to some stage below $s$. The set $G_s$ is grounded. Moreover, if $t \triangleleft s$, then $G_t \in G_s$.

Now, let $x$ be a non-empty set. Suppose that $x R s$. Let $y$ be the set of all $G_t$, for $t \triangleleft s$, such that there is a $z \in x$ such that $z R t$. Since any such $G_t$ is grounded, there is a stage $u$ such that $G_u \in y$ and $G_u$ is minimal. Since $G_u \in y$, there is a corresponding $z \in x$, such that $z R u$. So, $z$ must be minimal. Otherwise, there are $w \in z\cap x$ and a stage $v \triangleleft u$ such that $w R v$, hence $G_v \in y$. This contradicts the minimality of $G_u$, for $G_v \in G_u$.

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