Stuck on proof: If image of gauss map of connected n-surface is single point on the unit n-sphere, S must be a hyper plane 
The problem is from Thorpe's book Elementary Topics in Differential Geometry, where n-surfaces are described in terms of level sets of smooth functions with non-zero gradients on S.
My attempt, and where I am stuck -
Suppose the normal on S at all points is $n$. We need to show that then S is a hyperplane.
Now it is clear that for any point $p$ in S, the tangent space is a hyperplane at $p$ with normal $n$.
So let   $$H = \{x \in \mathbb{R}^{n+1}| (x - p) \cdot n  = 0\}$$
If I can show that for any other point $q$ in S, $q \in H$, then it is proved.
Since S is connected, there exists a continuous function $\alpha: [t_1, t_2] \longrightarrow S$ such that $\alpha(t_1) = p, \alpha(t_2) = q$, and $\alpha(t) \in S \; \forall \; t_1 \leq t \leq t_2$.
I am trying to show that $n\cdot p \neq n \cdot q$ leads to a contradiction. So I assume WLOG that $n \cdot p < n\cdot q$
By the IVT, I know that for every c between $n \cdot p, n\cdot q$, there exists $t$ so that $x = \alpha(t)$ satisfies $n \cdot x = c$.
Now if $$A = \{x \in \mathbb{R}^{n+1}| n \cdot p < n\cdot x < n \cdot q \}$$. 
Clearly, $A \subset S$, and I can see $A$ is open in S. 
According to the hint in the book, since S now contains an open set $A$, this is a contradiction. 
My question is- what is the contradiction here ?I am unable to close this out.
In one instance, I saw the suggestion that since  S contains an open set, its Gauss map must be the whole n-sphere, but  I do not see how this is either.
 A: This argument is way too complicated, and your claim that $A\subset S$ certainly does not follow; the set $A$ is an $(n+1)$-dimensional region, and $S$ is only an $n$-dimensional hypersurface. 
Consider the function $f(x) = (x-p)\cdot n$. Then for any curve $\alpha$ from $p$ to $q$ in $S$, defined on $[t_1,t_2]$ as you had it, we see that
$$(f\circ\alpha)'(t) = \alpha'(t)\cdot n = 0,$$
since $\alpha'(t)$ is in the tangent plane of $S$ at the point $\alpha(t)$. This means the function $f\circ\alpha$ is constant. Thus, $f\circ\alpha(t_2) = f\circ\alpha(t_1)=0$ and so $q\in H$, as well. (We use the fact that $S$ is connected to deduce that we can join $p$ to an arbitrary $q\in S$ by a path lying in $S$.)
EDIT: The best sense I can make of Thorpe's attempted argument is this. The reference to the vector field tells us that if any $p\in S$, an open subset of the hyperplane through $p$ with normal vector $v$ is contained in $S$. If we have a path $\alpha(t)$ in $S$, $t_1\le t\le t_2$, then this statement applies to every point of that path. For every $y$ value with $\alpha(t_1)\cdot v < y < \alpha(t_2)\cdot v$, we get a piece of a parallel hyperplane through some point $\alpha(t)$ on the curve. The union of these open subsets as we vary $t$ will fill up an open subset of $\Bbb R^{n+1}$, all of which is contained in $S$ [That contradicts the implicit function theorem and the fact that a level set must be $n$-dimensional.]. To be sure we actually get an open set here requires some analysis/topology, unless that differential equations result gives us existence for a uniform $\epsilon$ time independent of where we start.
