Principle of Transfinite Induction for On:
If $C\neq\varnothing, C\subseteq$ On then $\exists\alpha\in C\forall\beta\in C[\alpha\in\beta\vee\alpha=\beta]$
(1) As $C\neq\varnothing$, let $\alpha_0\in C$ .
(2) If for no $\beta\in C$ do we have $\beta\in\alpha_0$ then $\alpha_0$ was the $\in$-minimal element of $C$.
(3) Otherwise we have that $C\cap\alpha_0\neq\varnothing$. As $\alpha_0\in$ On, by definition $\in$ wellorders $\alpha_0$.
(4) Hence $C\cap\alpha_0$ is non-empty and so has an $\in$ minimal element $\alpha_1$; and then $\alpha_1$ is the minimal element of $C$.
I understand the general argument, but I am confused about $C\cap\alpha_0\neq\varnothing$ on line (3).
If $\alpha_0$ is the $\in$-minimal element of $C$ does that mean $C\cap\alpha_0=\varnothing$? If so, why?