Prove that if $C$ is an ordinal number and $r\in C$, then $r\cap C=\emptyset\Rightarrow r$ is the least element of $C$

Question

The definition of ordinal number is given by:

1. transitive: $\forall x\in S, x\subseteq S$

2. well-ordered:for every non-empty subset A ⊆ S, there exists an element m ∈ A such that m ≤ x ∀x ∈ A.

An ordinal number is a set that is transitive and is well-ordered by the relation $α < β ⇔ α ∈ β. $

Could someone prove that if $C$ is an ordinal number and $r\in C$, then $r\cap C=\emptyset\Rightarrow r$ is the least element of $C$ using this version of definition? Thanks in advance!

Suppose, in order to get a contradiction, that $r\cap C=\emptyset\land r$ is not the least element of $C$.

Then by definition of the least element, it means that it is not the case that $\forall c\in C, r\le c$.

Negating the expression gives that $(\exists c\in C )(r>c)$

By definition of $<$, we have $c\in r$.

As $r\in C$, $C$ is ordinal and hence transitive, we have $c\in C$.

Now we have both $c\in r$ and $c\in C$, thus the intersection of $r$ and $C$ is non-empty, a contradiction.