If $F$ is differentiable at $x_0$, $f$ is continuous at $x_0$
Solution 1:
The intuition is that if $f$ jumps at $x_0$, then the average of $f$ on intervals $(x_0 - h, x_0)$ to the left of $x_0$ will be smaller than the average on intervals $(x_0, x_0 + h)$ to the right of $x_0$.
Let $R = \inf_{x > x_0}f(x)$ be the right hand limit. Then $x > x_0 \implies f(x) \geq R$. Thus for $h > 0$, $\frac{1}{h}\int_{x_0}^{x_0 + h}f(t)\,dt \geq \frac{1}{h}Rh = R$. Get a similar estimate for $\frac{1}{h}\int_{x_0 - h}^{x_0}f(t)\,dt$ in terms of the left hand limit. Then let $h \to 0$.