Right versus left derivative by increasing one-sided derivatives

(I'll use the notation $f'_+$ and $f'_-$ for the right and left derivative.)

Unfortunately your proof does not work. This $$ \lim_{\varepsilon\to 0^+}\frac{f(z-\delta+\varepsilon)-f(z-\delta)}{\varepsilon} \overset{\varepsilon=\delta}{=} \lim_{\varepsilon\to 0^+}\frac{f(z)-f(z-\varepsilon)}{\varepsilon} $$ makes no sense because $\delta$ is fixed and the limits are taken for $\varepsilon \to 0 $. Also it would imply that $f'_+(z-\delta) = f'_-(z)$ for $0 < \delta < z-y$, which is true only if $f$ is linear between $y$ and $z$.

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What we can show is that $$ \tag{*} f'_+(y) \le \frac{f(z)-f(y)}{z-y} \le f'_-(z) $$ for $y < z$, which implies the desired conclusion. (This is motivated by the fact that both an increasing right derivative and an increasing left derivative imply that $f$ is convex.)

The proof of $(*)$ mimics the proof of Rolle's theorem and the mean-value theorem. We consider the function $$ g(x) = f(x) - (x-y)\frac{f(z)-f(y)}{z-y} $$ which is continuous and satisfies $g(y) = g(z)$. It follows that $g$ attains its maximum on the interval $[y, z]$ at some point $w \in [y, z)$. Then $g'_+(w) \le 0$ and it follows that $$ f'_+(y) \le f'_+(w) = g'_+(w) + \frac{f(z)-f(y)}{z-y} \le \frac{f(z)-f(y)}{z-y} \, . $$ This proves the left inequality in $(*)$, the proof of the right inequality works similarly.