Proving that Gauss map $N: S \rightarrow \mathbb{S}^{2}$ is surjective I'm taking a course in elementary differential geometry, and there was the following problem.

Let $S \subset \mathbb{R}^{3}$ be a non-empty, compact, and connected surface. Assume that (by the "Jordan-Brouwer" theorem) we can talk about "inside and outside" components of $S$. Let $N(p)$ be the outward normal vector at $p \in S$. Prove that the Gauss map $N: S \rightarrow \mathbb{S}^{2}$ is surjective.

Solution: Fix a unit vector $v \in \mathbb{S}^{2},$ and consider the family of planes
$$
P_{c}=\left\{x \in \mathbb{R}^{3} \mid\langle x, v\rangle=c\right\}, \quad c \in \mathbb{R}
$$
Since $S$ is compact, the function $x \mapsto\langle x, v\rangle$ on $S$ achieves its maximum value at some point $p \in S$. Define $c_{0}=\langle p, v\rangle$. Then $S \cap P_{c}$ is empty for all $c>c_{0}$ whereas $S \cap P_{c_{0}}$ is not empty. We claim that $v=N(p)$. To see this, consider an arbitrary smooth curve $\alpha:(-\epsilon, \epsilon) \rightarrow S$, with $\alpha(0)=p$ and $\alpha^{\prime}(0)=w \in T_{p} S$. Since $f(t)=\langle\alpha(t), v\rangle$ is maximised at $t=0$, we must have $0=f^{\prime}(0)=\langle w, v\rangle$. Therefore, $v$ is orthogonal to any tangent vector to $S$ at $p$. In fact, $v$ must be the outward unit normal vector, since all of $S$ lies in the half-space $\langle x, v\rangle \leq\langle p, v\rangle$. As $v \in \mathbb{S}^{2}$ was arbitrary, the Gauss map $N: S \rightarrow \mathbb{S}^{2}$ must be surjective.
I don't understand the line "In fact, $v$ must be the outward unit normal vector, since all of $S$ lies in the half-space $\langle x, v\rangle \leq\langle p, v\rangle$." Why is this? I can't understand this intuitively, nor can I understand it formally (with an actual proof). Does anyone know why this is?
 A: The catch lies in this part of the problem:

Assume that (by the "Jordan-Brouwer" theorem) we can talk about "inside and outside" components of $S$.

Note that this wasn't used elsewhere in the presented reasoning.

For an (at least) intuitive explanation, note that $P_{c_0}$ divides the whole space into two half-spaces:
$$
P^+_{c_0}=\left\{x \in \mathbb{R}^{3} \mid\langle x, v\rangle > c\right\}, 
\quad 
P^-_{c_0}=\left\{x \in \mathbb{R}^{3} \mid\langle x, v\rangle \le c\right\}.
$$
According to the proof, all $S$ is contained in $P^-_{c_0}$, since $c_0$ was chosen as the maximal value of $\langle x, v \rangle$. Hence, the whole half-space $P^+_{c_0}$ is on the outside of $S$.
On the other hand, the vector $v$ (think of it as anchored at $p \in P_{c_0}$) points towards $P^+_{c_0}$, e.g. in the sense that
$$
\langle p+tv, v \rangle = c_0 + t > c_0 
\quad \text{for } t>0,
$$
so each point $p+tv$ (with positive $t$) lies in $P^+_{c_0}$. This should explain why we call it the outer normal.
