Proof of property of normals to a star shaped region ("elementary" vector calculus lemma in Evans' PDE) I cannot follow the proof of the following Lemma (Evans's PDE, p. 515).

Lemma. If $U\subset\mathbb{R}^n$ is an open star-shaped region with $\partial U\in C^1$, then $x\cdot\nu(x)\ge0$ $\forall x\in \partial U$, where $\nu$ denotes the outward normal.

Proof.
We assume $U$ is star-shaped with respect to the origin $0$.
Since $\partial U$ is $C^1$, if $x\in\partial U$ then for each $\varepsilon >0$ there exists $\delta >0$ such that $|y-x|<\delta$ and $y\in\bar{U}$ imply 
$$\nu(x)\cdot\frac{y-x}{|y-x|}\le\varepsilon.$$

Why is that?

In particular $$\limsup_{y \to x, y \in \overline{U}} \nu(x) \cdot \frac{y-x}{|y-x|} \le 0.$$
Let $y = \lambda x$ for $0 < \lambda < 1$. Then $y \in \overline{U}$ since $U$ is star-shaped  with respect to the origin $0$. Therefore $$\nu(x)\cdot\frac{x}{|x|} = - \lim_{\lambda \to 1^-} \nu(x) \cdot\frac{\lambda x-x}{|\lambda x-x|}.$$

Why is that?

But we have shown that the last term is $\ge 0$, hence we are done.
 A: I'll sketch answers to each of your "Why is that?"s and then give some intuition as to why the result should be true, without giving a full proof.
To your first "Why is that?" --- The fact that $\partial U$ is $C^1$ means that the normal vector field is continuous and that every point on the boundary has a tangent plane. Let's interpret the equation in question: For a point $y$ in $U$ or on its boundary sufficiently close to $x$, the unit vector from $x$ to $y$ is "almost pointing behind" $\nu(x)$.
To precisely prove this, write $U = F^{-1}(-\infty,0)$ and $\partial U = F^{-1}(0)$ and $\nu = \nabla F$ for a $C^1$ function $F$. Interpreted this way, the claim is: for any $\epsilon > 0$, there is a $\delta > 0$ such that if $|y-x|<\delta$ and $F(y) \leq F(x)$, then
$$ \nabla F(x) \cdot \frac{y-x}{|y-x|} < \epsilon $$
To your second "Why is that?" --- just some algebra: 
$$ \frac{\lambda x - x}{|\lambda x - x|} = \frac{(\lambda - 1)x}{|\lambda - 1||x|} 
 = -\frac{x}{|x|}$$
as $\lambda \in (0,1)$.
What's the intuition?
The sign of the inner product measures whether one vector points in the direction of another vector. If $x\cdot y > 0$ then $x$ and $y$ are pointed in the same direction; if $x\cdot y < 0$, then $x$ and $y$ are pointed in opposite directions.
In the case of a star-shaped domain $U$, for any $x\in \partial U$, the lemma avers $x$ and $\nu(x)$ are always pointed in the same direction. In other words, $\nu(x)$ can never point back toward $x$. Why should this be true? If $\nu(x)$ pointed outward and back toward $x$, then as we approached $x$ from $0$, we would be outside the domain; hence $U$ could not be star-shaped with respect to $0$ after all.
The proof in some sense is backwards. In proving this lemma from scratch, one should start with this intuition, write down the condition, then be led to conduct the analysis to your first "Why is that?"
