In my humble opinion, the general principle should have a very similar proof to the original principle to be considered a proper generalization. After reconstructing a proof of your original principle, I obtained the following generalization.

Theorem. Let $X$ be a poset in which every nonempty set has an inf. (Note that $X$ has a minimal element $0$ and that every set which is bounded above has a sup.) If $S \subseteq X$ is such that:

1. If $x \in S$ and $y < x$ then $y \in S$.
2. If $x \in S$ and $x < y$ then there is some $z \in S$ such that $x < z \leq y$.
3. Every subset of $S$ which is bounded above (in $X$) has its sup in $S$.

Then $S = X$. (Note that 3 formally implies that $0 \in S$ since $0$ is the sup of the empty set.)

This is slightly different than the original principle, but one recovers the original principle by applying the above theorem to the subset $S' = \{x \in X : [0,x] \subseteq S\}$ instead of $S$.

After writing down the proof I had in mind, I realized that condition 1 is not necessary.

Proof of Theorem. Suppose $y \notin S$. Let $x = \sup\{z \in S : z \leq y\}$. By 3, $x \in S$ and so $x < y$. By 2, there is a $z \in S$ such that $x < z \leq y$, which contradicts the definition of $x$. QED

1

In my humble opinion, the general principle should have a very similar proof to the original principle to be considered a proper generalization. After reconstructing a proof of your original principle, I obtained the following generalization.

Theorem. Let $X$ be a poset in which every nonempty set has an inf. (Note that $X$ has a minimal element $0$ and that every set which is bounded above has a sup.) If $S \subseteq X$ is such that:

1. If $x \in S$ and $y < x$ then $y \in S$.
2. If $x \in S$ and $x < y$ then there is some $z \in S$ such that $x < z \leq y$.
3. Every subset of $S$ which is bounded above (in $X$) has its sup in $S$.

Then $S = X$. (Note that 3 formally implies that $0 \in S$ since $0$ is the sup of the empty set.)

This is slightly different than the original principle, but one recovers the original principle by applying the above theorem to the subset $S' = \{x \in X : [0,x] \subseteq S\}$ instead of $S$.