Steps: 

 1. Consider $\phi_r(x)$ to be $1$ on $[x-r,x+r]$ and $0$ outside of $[x-2r,x+2r]$ and connects linearly from $x-2r$ to $x-r$ and from $x+r$ to $x+2r$. (I know this is not exactly $C_c^\infty$ but it is easier to work with and one can always choose the "smoothed out" version.)
 2. $\phi_r'(x)=\frac{1}{r}$ on $[x-2r,x-r]$ and $-\frac{1}{r}$ on $[x+r,x+2r]$. Heuristically, this converges to Dirac delta from above and below, i.e. $\delta_{x^-}-\delta_{x^+}$
 3. $Du(\{x\})=\lim_{r\to0}\int_I\phi_r(x)d(Du)(x)=\lim_{r\to0}\int_I\phi'_r(x)u(x)dx=u(x^-)-u(x^+)$
 4. By definition, $\forall \epsilon >0,\ \exists R>0$ s.t. $\forall r<R$ (expect for Lebesgue measure $0$ set), $|u(x-r)-u(x^-)|<\epsilon$. Similarly for the plus part with the same $R$!
 5. Combine the definition of Lebesgue point, and $|u(x)-u(x^\pm)|\leq |u(x^\pm)-u(y)|+|u(y)-u(x)|$ to get

$$\frac{1}{2r}\int_{x-r}^{x+r}|u(y)-u(x)|dy\geq \frac{1}{2}(|u(x)-u(x^-)|-\epsilon)+\frac{1}{2}(|u(x)-u(x^+)|-\epsilon)\geq |u(x^-)-u(x^+)|-2\epsilon$$

**Remark:** I am not so fond of what I did in Step 2 and 3.