An ordinal is defined as a woset $(X, ≤)$ such that $X_a = a$ for all $a ∈ X$.
How can we prove that a limit ordinal is still well-founded? I don't see why there could not be an infinite descending chain of the inclusion or subset relation in a limit ordinal. For example $\omega$ or any other limit ordinal.
Thanks.
On
As I suggested in the comments above, this is true for any woset--ordinal or otherwise.
Suppose that $\langle X,<\rangle$ is a woset, and that $\langle a_n\rangle_{n=0}^\infty$ is a sequence of points of $X$ such that $a_{n+1}\le a_n$ for all $n.$ Now, what can you say about the subset $\{a_n:n\ge0\}$ of $X$? What does that tell you about the sequence?
Suppose $\alpha$ is a limit ordinal and $\alpha>\beta_1>\beta_2>\beta_3>...$ is a descending sequence in $\alpha$. Then think about the first term of this sequence, $\beta_1$. Since everything below $\alpha$ is an ordinal (since $\alpha$ is an ordinal), $\beta_1$ is itself an ordinal. But then $\beta_2>\beta_3>\beta_4>...$ is a descending sequence through $\beta_1$!
What this really shows is that any set of ordinals is well-ordered by $\in$.
You mention the subset relation in the context of ordinals. This is the wrong relation to have in mind! There are indeed descending subset sequences in $\omega$: e.g. $$\omega\supset (\omega\setminus 1)\supset(\omega\setminus 2)\supset(\omega\setminus 3)\supset . . .$$ But note that the sets in this sequence aren't ordinals - in particular, they're not elements of $\omega$!