Your axiom schema is precisely the statement "all sets are of finite rank", that isequivalent to being an $V=V_\omega$$\omega$-model.
Let $X$ be any set and let $\phi(x)$ beWorking inside an $x=x$$\omega$-model, byany counter example to your schema will be counter example of the axiom there exists someof foundation of ZFC, as you can use $m\in\omega$ such that there is no$\le$ and the given $m$$\phi$ of the counter-descendingexample to construct the counter-example sequence starting from $X$, so there is no $m-1$-descensingthis sequence starting from any $y\in x$is of infinite length externally, and so on and so forthbut the model has the same finite numbers, so you get that the rank of $X$it is less than $m$ ($X\in V_m$)infinite internally as well.
Now ifassume the model is not an $X\in V_\omega$ we know that$\omega$-model, let $X\in V_m$ for some$\phi(x)$ be $m\in\omega$, so your schema$x=x$ and $X$ be a $k$-nested singleton where $k$ is obviously correct.
(Note that I didn't usesome non standard natural, then $X, \phi$ will be a counter example to the well ordering onschema as the universe)sequence defined as $x_{i+1}=\min_\le\{x\in x_i\}, x_0=X$ won't stop in any standard step.