Ordinals are defined as a set which is transitive and all its elements are transitive.
In the proof of linearity of ordinals i.e. $$\forall a \forall b : a<b \vee a>b \vee a=b $$
We assume that this fails and choose minimal $a$ such that $b$ is neither less than or greater than or equal to $a$. How do we know we can choose such a minimal element? i.e.
1) How do we know the set of such $a$ has all elements comparable (aren't we proving this very thing ?)?
2) If this set has all elements comparable then how is a minimal element guaranteed?
One can go further and ask does 1) makes sense, i.e. do such $a$s make a set. How does one answer this question?
I'd suggest a different proof. Let $\alpha, \beta$ be ordinals. If $\alpha = \beta$, then there is nothing do to. So, suppose wlog that $\beta \not \subseteq \alpha$. Let $\gamma = \min_{\in} \beta \setminus \alpha$. (Such a minimum exists by regularity and the transitivity of $\beta$.)
Now show that $\gamma = \alpha$ (using extensionality) and hence $\alpha \in \beta$.