Observe that for given ordinals $A$ and $B$ the following claims hold true:
- If $A$ is a proper subset of $B$, then $A \in B$.
- $A \cap B$ is an ordinal.
Now, given ordinals $A \neq B$ consider $A \cap B \subseteq A,B$. If $A \cap B = A$, then $A \subseteq B$, so that either $A = B$ or $A \in B$. Analog $A \cap B = B$ implies either $B = A$ or $B \in A$. If $A \cap B$ is a proper subset of both $A$ and $B$, then $A \cap B \in A \cap B$, which contradicts the axiom of regularity.
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.
Best Answer
We don't need to know that something is a set in order to order it.
Formally speaking if $A$ is a class we can define an order on $A$ to be another class $R$ such that $z\in R$ means that $z=\langle x,y\rangle$ and $x,y\in A$, and moreover $R$ satisfies the axioms of an ordered set (whichever you prefer them to be, irreflexive and transitive; or reflexive, antisymmetric and transitive).
Now we can say that $\in\upharpoonright\sf Ord\times Ord$ is a well-ordering of $\sf Ord$.
But if we assume that $\sf Ord$ is a set then we can show that it is in fact an ordinal itself, therefore $\sf Ord\in Ord$. However this means that $\sf Ord$ is not well-founded, because $\{\sf Ord\}$ is a non-empty subset without a minimal element!