I have two $C^1$ functions $f$ and $g$ which go from $\mathbb{R}^{N-1}\rightarrow\mathbb{R}$. For some $k_1,k_2\in\mathbb{R}^N$, I know that locally about a point $x_0\in\mathbb{R}^N$ the sets;
$$F=\{\vec{x}\in\mathbb{R}^N \text{ where }\vec{x}=(x_1,...x_{k_1-1},f(\vec{x}_{|k_1}),x_{k_1+1},...x_n) \text{ for some } \vec{x}_{|k_1}=(x_1,...x_{k_1-1},x_{k_1+1},...x_N)\in\mathbb{R}^{N-1}\}$$ $$G=\{\vec{x}\in\mathbb{R}^N \text{ where }\vec{x}=(x_1,...x_{k_2-1},g(\vec{x}_{|k_1}),x_{k_2+1},...x_n) \text{ for some } \vec{x}_{|k_2}=(x_1,...x_{k_2-1},x_{k_2+1},...x_N)\in\mathbb{R}^{N-1}\}$$
are continuously differentiable curves/surfaces. Let $B(x_0,\epsilon)$ be an $\epsilon$-ball about $x_0$. I want to show that for $\epsilon$ small enough, the set;
$$(F\cup G) \cap B(x_0,\epsilon)$$
partitions $B(x_0,\epsilon)$ into at most $4$ disjoint path-connected sets. I believe this is true but am really struggling to formulate it.
I tried using implicit function theorem on $f(\vec{x}_{|k_1})-g(\vec{x}_{|k_2})$ but I don't think I can as $y=f(\vec{x}_{|k_1})$ and $y=g(\vec{x}_{|k_2})$ could overlap in a neighbourhood of $x_0$ meaning $f(\vec{x}_{|k_1})-g(\vec{x}_{|k_2})$ could equal zero in a neighbourhood of $x_0$ and therefore have $0$ gradient. How can I show this?