I feel like the problem here is stated incorrectly:
I think we need $F$ to not just be (Frechet) differentiable but also continuously differentiable, as in the map that maps each $x \in O \to F'(x)$ itself should be a continuous map because that is what was stated in the inverse function theorem. Given how the problem is currently stated, we can't apply it to show that $F(O)$ is open.
I wouldn't know about Banach spaces, but if $X=Y=\Bbb R$ then just differentiability is enough. Say $O\subset \Bbb R$ is open. Then $O=\bigcup I_j$ for a collection of open intervals $I_j$. It's enough to show $F(I_j)$ is open. Since $F$ is continuous, $F(I_j)$ is some sort of interval.
Suppose that $F(I_j)=(a,b]$ or $F(I_j)=[a,b]$. Choose $p\in I_j$ with $F(p)=b$. Now, since $I_j$ is open, $F(p)=\max_{x\in I_j} F(x)$ shows that $F'(p)=0$, contradiction. Similarly for $[a,b)$; hence $I_j=(a,b)$.