Smooth case of Jordan Brouwer Separation Theorem; proving there is more than 1 connected component

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Im having trouble proving something. I am using smooth degree theory to prove the Jordan Brouwer Separation Theorem for a smooth compact connected oriented $n$ dimensional surface $M \subset \mathbb{R}^{n + 1}$. Let $\Omega = \mathbb{R}^{n + 1} \setminus M$. For each $p \in \Omega$ define a map $F_p \colon M \to S^n$ by $$F_p(x) = \frac{x - p}{||x - p||}.$$ I have shown that the degree of $F_p$ is constant on each connected component of $\Omega$. In order to prove that there is more than one connected component, it suffices to show that there are two points $p_0, p_1 \in \Omega$ such that the degrees of $F_{p_0}$ and $F_{p_1}$ differ. If I can find a point $p \in \Omega$ such that $F_p$ has a regular value with nonempty inverse image, then the rest of the proof is easy, but I am not able to prove that such a $p \in \Omega$ exists. Any help on proving (or disproving) this is appreciated.

I think this is equivalent to showing that there exists a ray $r$ that is never tangent to $M$ and intersects $M$ at least once.

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Since $M$ is flat, locally (in the neighborhood of some point $x_0$) it looks like a hyperplane. From the local picture it is clear all non-tangent at $x_0$ directions are regular values.