The connectivity of the intersection of hypersurface and ball $u$ is a function defined on a connected open set $\Omega$ of $\mathbb R^n$ containing $0$ such that $u \in C^2(\Omega)$ and $u(0)=0$. Consider the hypersurface $X=\{(x,u(x))~|~x\in\Omega\}$ and the ball $B_r(0)=\{ x \in \mathbb R^{n+1}~|~|x|<r \}$,then can we find a small $r>0$ such that $X \cap B_r(0)$ is connected?
 A: Yes, in fact $u\in C^1$ is enough. First, consider the case when $\nabla u(0)=0$. Let $v(x)=|x|^2+u(x)^2$. It suffices to prove that for sufficiently small $r>0$ the set $Y_r=\{x:v(x)<r^2\}$ is connected, because $X\cap B_r$ is the image of $Y_r$ under continuous map $x\mapsto (x,u(x))$. 
Clearly $Y_r$ is open. Suppose it has a connected component $W$ that does not contain $0$. Since $v=r^2$ on $\partial W$, it follows that $W$ contains a critical point of $v$. But
$$|\nabla v(x)| = |2x+2u(x)\nabla u(x)|\ge 2|x|-2|u(x)|\,|\nabla u(x)| \tag1 $$
where the second term is much smaller than the first: $u(x)=O(|x|^2)$ and $|\nabla u(x)|=O(|x|)$. 

In the general case, observe that $X$  is a $C^1$-smooth embedded surface, and therefore has a tangent plane at $0$. Choose an orthogonal system of coordinates $(y_j)$ so that $y_1,\dots,y_n$ lie in the tangent plane. Then $X$ is locally represented as $y_{n+1}=\tilde u(y_1,\dots,y_n)$ where the function $\tilde u$ satisfies $\tilde u(0)=0$ and $\nabla \tilde u(0)=0$. The special case above applies: the  connectedness of $X\cap B_r(0)$ is not affected by passing from one coordinate system to another. 
