 Solution

Use transfinite induction. For any $\beta <\alpha$, the induction hypothesis and strict increase give $\beta \le F(\beta )<F(\alpha )$. Thus every $\beta <\alpha$ belongs to $F(\alpha )$, so $\alpha \subseteq F(\alpha )$, i.e.This includes $\alpha =0$ and does not yet use continuity of the normal function on an ordinal.