First note that since $\mathsf{ZFC}$ is just a formal first-order theory, it doesn't pick out exactly what form a model's interpretation of the $\in$ symbol takes. So there are models of $\mathsf{ZFC}$ where the $\in$ relation is something other than the real membership relation. Actually, this point is crucial in the development of Boolean-valued models. In such circumstances, what $M$ thinks of as an ordinal will not be what $\mathbf{V}$ thinks of as an ordinal.
But I assume that you are thinking of models where $\in$ is interpreted as the real membership relation. But we do have the following facts.
- If $\kappa$ is an inaccessible cardinal, then $V_\kappa \models \mathsf{ZFC}$, and $V_\kappa$ clearly does not contain all ordinals.
- Since the underlying language of $\mathsf{ZFC}$ is countable, by the Löwenheim–Skolem Theorem (if it is consistent) $\mathsf{ZFC}$ has countable models. Taking any countable model $\mathfrak{M} = ( M , E )$ of $\mathsf{ZFC}$ there must be ordinals of $\mathbf{V}$ which are not in $M$. It might be that what $\mathfrak{M}$ thinks of as ordinals are not the ordinals of $\mathbf{V}$, but even if we are "lucky" $\mathfrak{M}$'s version of ordinals is "correct" (which will occur if $M$ is a transitive set and $E$ is the real membership relation, see below) there must be ordinals of $\mathbf{V}$ not appearing in $M$.
In the second point above I mentioned transitive set models of $\mathsf{ZFC}$ (where $\in$ is interpreted as real membership). The existence of models of this kind is an even stronger assumption than the simple consistency of $\mathsf{ZFC}$, then we cannot show from $\mathsf{ZFC}$ (or even $\mathsf{ZFC}+\mathrm{Con}(\mathsf{ZFC})$) that such exist.
But transitive (possibly proper class) models of $\mathsf{ZFC}$ must agree with $\mathbf{V}$ about what ordinals are (if not "how many" there are):
If $M$ is a transitive model of $\mathsf{ZFC}$, and $a \in M$, then $M \models \text{"}a\text{ is an ordinal"}$ iff $a$ really is an ordinal (in $\mathbf{V}$).
This is due to the fact that there is a $\Delta_0$-formula $\phi(x)$ (i.e., a formula in which all quantifiers can be expressed in the form $(\exists x \in y)$ for $(\forall x \in y)$) which in $\mathsf{ZFC}$ defines the property of being an ordinal, and it is known that $\Delta_0$-formulas are absolute for transitive models.
A transitive model $\mathbf{M}$ of $\mathsf{ZFC}$ which contains all the ordinals (from $\mathbf{V}$) is called an inner model.
The notions are very different, to my best understanding.
As far as I could understand it, the pre-Cantorian infinity was primarily a notion of a length which is longer than any other. It was closer to the infinity we meet in real analysis, rather than the infinity we deal with in set theory. Despite the informal similarities that one can talk about "$f(\infty)$" and "$\alpha\in\sf Ord$", as if both these were actual objects of their universes.
Cantor noticed that one can consider a queue of infinite length, and that it makes sense to have an infinite queue, and then some. From there he defined the ordinals and the cardinals, and the rest is history.
On the other hand, from a modern point of view, the class of ordinals need not be absolute. One can consider end-extensions, which really add more ordinals to the universe, a typical example is taking an inaccessible $\kappa$, then $\bf V$ is an end-extension of $V_\kappa$.
In one considers theories like the Tarski-Grothendieck, or equivalently the theory $\sf ZFC+$"There is a proper class of inaccessible cardinals" then one can consider the least inaccessible as the universe of sets, and then one can always find a larger and larger universe with more and more ordinals. If one goes to even stronger cardinals (e.g. Woodin cardinals) then these properties of "extending higher and higher" can get stronger and stronger.
From another point of view, one can consider the multiverse approach which was proposed by Joel D. Hamkins (see J. D. Hamkins, The set-theoretic multiverse, Review of Symbolic Logic 5 (2012), 416–449.), which says among other things that every universe of set theory is really a countable model from the point of view of another universe (see also this blog post by Francois Dorais).
This is a much stronger assumption that the proper class of inaccessible cardinals. There we had a sequence of universes, each larger than the last, but none was countable in any of the extensions. In fact they all agreed on their common cardinalities and sets. In this case the universes become smaller and smaller as we go along.
So whereas the potential infinity of the early 19th century was somewhat of "the energy of an unstoppable object"; the class of ordinals is something vastly more frightening in size. But at the same time, the calculus notion of infinity is very coarse and hardly at all manageable. On the other hand, the class of ordinals is a concrete class (for a given universe, of course), which can be managed and manipulated internally. This is added by the above facts that the ordinals can be made a set of a larger universe; or even a countable set of a much larger universe.
Best Answer
There is no such property--the proof that $\mathbf{ORD}$ is well-ordered has nothing to do with it being a set. Indeed, you can prove in ZF that $\mathbf{ORD}$ is a well-ordered class, meaning that every nonempty subset of $\mathbf{ORD}$ has a least element. Namely, given a nonempty subset $X\subseteq\mathbf{ORD}$, pick an element $\alpha\in X$. If $\alpha$ is the least element of $X$, we're done. Otherwise, $\alpha\cap X$ is nonempty, and so has a least element $\beta$ since $\alpha$ is an ordinal. Then $\beta$ is also the least element of $X$, since any $\gamma\in X\setminus\alpha$ must be greater than or equal to $\alpha$ and thus greater than $\beta$. (Here I assume you already know that the ordinals are totally ordered by $\in$, which takes some work to prove.)
The relevance of $\mathbf{ORD}$ being a set is that $\mathbf{ORD}$ is by definition the class of all sets that are transitive and well-ordered by $\in$. So, to conclude that $\mathbf{ORD}\in\mathbf{ORD}$, you need to know that $\mathbf{ORD}$ is a set.
Note that essentially the same argument can be made in other contexts where everything is a set, which you may find less confusing. For instance, if you define $\omega$ as the set of finite ordinals, then you can prove by this argument that $\omega$ must be infinite as follows. First, you prove that $\omega$ is an ordinal. Then, if $\omega$ were finite, you conclude that $\omega$ would be a finite ordinal and hence an element of $\omega$. But then $\{\omega\}$ would be a subset of $\omega$ with no least element, which is a contradiction.