A different proof of Urysohn's Lemma I want to show Urysohn's Lemma following this approach.
Let $K\subset V \subset \mathbb R^N$, with $K$ compact  and $V$ open. Set 
$\delta= \mbox{ dist } (K,V^c)$ and take $\epsilon<\delta/2$. Define $V_\epsilon=\{x\in \mathbb R^N : 
\mbox{ dist } (x,K) < \epsilon\}$ and consider the following function
$$
g(x)=\frac{|V_\epsilon\cap B_\epsilon(x)|}{|B_\epsilon|}, \quad \mbox {where } B_\epsilon(x) \mbox{ denotes the ball centered at } x \mbox { and radius } \epsilon.
$$ 
Need to show that $g\in \mathcal C_c(\mathbb R^N)$ and that $\forall x \in \mathbb R^N$ we have  $\chi_K(x)\le g(x)\le \chi_{V}(x)$. 
In fact I don't know how to proceed.
 A: All claims except the continuity of $g$ are clear. For this, suppose $x_n\to x$, so 
$$
m(B_\epsilon) \lvert g(x_n) - g(x) \rvert = \lvert m( V_\epsilon \cap B_\epsilon(x_n)) - m(V_\epsilon \cap B_\epsilon(x)) \rvert \le m( V_\epsilon \cap (B_\epsilon(x_n) \mathbin{\square} B_\epsilon(x) ) ) \le 
m(B_\epsilon(x_n) \mathbin{\square} B_\epsilon(x)) =
m(B_\epsilon(x_n)) + m( B_\epsilon(x)) -2
m(B_\epsilon(x_n) \cap B_\epsilon(x)) = 2 (m( B_\epsilon(x)) - m(B_\epsilon(x_n) \cap B_\epsilon(x))
$$
Here I have written $\square$ for symmetric difference and used the inclusion exclusion principle. Hence the result will follow if we can prove that $m(B \cap (B+x)) \to m(B)$ as $x\to 0$. This follows from this result.
A: You got to a point (in your Oct 20 at 9:37 comment) when you need to show that $|B_\epsilon(x_n)) \setminus B_\epsilon(x))|\to0$ as $x_n\to x$. But, I don't quite see how you came to 
$|B_\epsilon(x_n)) \setminus B_\epsilon(x))|$.
(I think it is right, but the steps shown are not clear).
You started your comment with 
$g(x_n)-g(x)=\frac{1}{|B_\epsilon|} |(V_\epsilon \cap B_\epsilon(x_n))-(V_\epsilon \cap B_\epsilon(x))|$ and I would think that one should instead write 
$g(x_n)-g(x)=\frac{1}{|B_\epsilon|}\Bigl(|(V_\epsilon \cap B_\epsilon(x_n))|-|(V_\epsilon \cap B_\epsilon(x))|\Bigl)$. In addition to this correction, the LHS should be $|g(x_n)-g(x)|$, so $|g(x_n)-g(x)|=\frac{1}{|B_\epsilon|}|\Bigl(|(V_\epsilon \cap B_\epsilon(x_n))|-|(V_\epsilon \cap B_\epsilon(x))|\Bigl)|$ (where some of the vertical bars denote measure, and some absolute value). 
Ultimately you need to show that when $y$ is close to $x$ then the measure 
$|V_\epsilon \cap B_\epsilon(y)|$ is close to the measure $|V_\epsilon \cap B_\epsilon(x)|$. 
Consider the case when $|V_\epsilon \cap B_\epsilon(y)|\ge|V_\epsilon \cap B_\epsilon(x)|$ (the other case being symmetric to this one). 
Then $0\le|V_\epsilon \cap B_\epsilon(y)|-|V_\epsilon \cap B_\epsilon(x)|\le$
$\le|V_\epsilon \cap B_\epsilon(y)|-|(V_\epsilon \cap B_\epsilon(y))\cap(V_\epsilon \cap B_\epsilon(x))|=$
$=|(V_\epsilon \cap B_\epsilon(y))\setminus(V_\epsilon \cap B_\epsilon(x))|=$
$=|V_\epsilon \cap (B_\epsilon(y)\setminus B_\epsilon(x))|\le$
$\le|B_\epsilon(y)\setminus B_\epsilon(x)|$. 
Now, it should be obvious that $|B_\epsilon(y)\setminus B_\epsilon(x)|\to0$ 
as $y\to x$. I believe this might have been what @DanielFischer had in mind when he suggested making a sketch, in an Oct 19 at 12:45 comment. A picture in the plane is easy (and in $\Bbb R^3$ also easy enough). In the plane, the two balls (or disks) mostly overlap, except for two small "moons" (or crescents). The measure of these moons goes to $0$ (as $y\to x$). The formula for the volume $n$-dimensional ball is somewhat messy (if you would rather not rely on a picture), but it would be easier to come up with an estimate for the volume of each crescent, and one such estimate is the product of the norm $|x-y|$ with the "area" of the $(n-1)$-dimensional "disk". (I am using Cavalieri's principle for this estimate.) The area of $(n-1)$-dimensional disk in turn does not exceed the "area" of $(n-1)$-dimensional square with side-length the same as the diameter of $B_\epsilon$ (which is $2\epsilon$). 
Thus the area of such $(n-1)$-dimensional square with side-length $2\epsilon$ is 
$(2\epsilon)^{n-1}=2^{n-1}\epsilon^{n-1}$. 
The estimate for the volume of each crescent is $2^{n-1}\epsilon^{n-1}|x-y|$, 
and we have two crescent so the total volume estimate is 
$2^n\epsilon^{n-1}|x-y|$. Since $n$ and $\epsilon$ are fixed, we have that 
$2^n\epsilon^{n-1}|x-y|\to0$ as $y\to x$. (You may want to verify, and if necessary correct my estimates, to confirm that this approach works.) 
Again, it helps to draw a picture. 
