Principle of Transfinite Induction for On:
If $C\neq\varnothing, C\subseteq$ On then $\exists\alpha\in C\forall\beta\in C[\alpha\in\beta\vee\alpha=\beta]$
Proof:
(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?
Note that (3) begins with "Otherwise", meaning if $\alpha_0$ wasn't the minimal element of $C$, then $C\cap\alpha_0\neq\varnothing$ and in which case $\alpha_1=\min\alpha_0\cap C$ is also the minimal ordinal in $C$.