The ordinals are totally ordered, i.e. for $\alpha,\beta\in\mathrm{Ord}$ we have
$$\alpha\in\beta\quad\vee\quad\beta\in\alpha\quad\vee\quad\alpha=\beta.$$
Each proof I have seen so far started like this: Assume ther are incomparable ordinals. Choose an $\in$-minimal ordinal $\alpha$ so that there is an incomparable ordinal $\beta$.
But how can we do this? How do I know that there is an $\in$-minimal ordinal? I think the axiom of regularity does not help me here, because $\mathrm{Ord}$ is a proper class. Do I have to show that $\mathrm{Ord}$ is well-founded before proving totallity?
You can prove that given any ordinal $\alpha$, the ordinals below $\alpha$ are linearly ordered.
Now if $\alpha$ and $\beta$ are two ordinals, consider $x$ to be the transitive closure of $\{\alpha,\beta\}$. I claim that $x$ is a transitive set (by definition) whose elements are all transitive sets, this is because $$x=\operatorname{tcl}(\{\alpha\})\cup\operatorname{tcl}(\{\beta\})=\alpha\cup\beta\cup\{\alpha,\beta\}.$$
So $x$ is now an ordinal such that both $\alpha$ and $\beta$ are less than $x$. So they are comparable by the general proof given before.