$f$ is increasing if and only if $f_{-}'\leq f_{+}'$

Solution 1:

Neither of the implications is true. In fact:

$$f:(0,1)\to \mathbb{R}$$ $$x\mapsto -x$$

This map is strictly decreasing but it satisfies $$f'_{-}(x)\leq f'_{+}(x)$$(in fact they are always equal).

Now consider:

$$g(x)=\begin{cases} x \text{ if }0\leq x\leq\frac12\\ \frac12 \text{ if }\frac12 \leq x \leq 1 \end{cases}$$

This function is increasing, but it doesn't satisfy your condition in $x=\frac12$.

Solution 2:

Suppose the hypothesis holds. Then $f(x_1)\leq f(x_2)\leq f(x_3)$ for alle $x_1<x_2<x_3$ in $(a,b)$, and so the quotients hold $$ \frac{f(x_2)-f(x_1)}{x_2-x_1}\leq \frac{f(x_3)-f(x_1)}{x_3-x_1} $$

This is not true, take for instance $f(x)=4x-x^2$ increasing on the interval $[0,2]$

$\begin{array}{|l}f(0)=0\\f(1)=3\\f(2)=4\end{array}\quad $ but $\quad \overbrace{\dfrac{f(1)-f(0)}{1-0}}^3>\overbrace{\dfrac{f(2)-f(0)}{2-0}}^2$

Therefore the remaining of your proof is invalidated.