Open neighborhood on proof of implicit function theorem.

Question

Consider the $C^1$ function $F:\mathbb{R}^n\times\mathbb{R}^m\rightarrow\mathbb{R}^+$ with $F(0,0)=0$, the set $S_r(0):=\lbrace x\in\mathbb{R}^m : ||x||=r \rbrace$ and $\alpha:=\min_{y\in S_r(0)}F(0,y)$. For every point $y\in S_r(0)$ the pre-image of the interval $]\frac{\alpha}{2};\infty[$ of the continuous function $g_y(x):=F(x,y)$ is an open neighborhood of $0\in\mathbb{R}^n$, say $\mathcal{O}_y$. Am I guaranteed that there exist an open neighborhood $U$ of $0\in\mathbb{R}^n$ such that $F(x,y)\in ]\frac{\alpha}{2};\infty[,\ \forall x\in U, \forall y\in S_r(0)$?. How do I obtain this neighborhood? because if I take the intersection $U:=\bigcap_{y\in S_{r}(0)} \mathcal{O}_{y}$ it could be a closed set or worse just the point $0$. The context of this is the proof of the implicit function theorem where the neighborhood is obtain with help of $F(x,y)=||f(x,y)||$ and the implicit function is proved for $f$.