Uniform continuity of difference function

Question

Assume that both $f$ and $f'$ are uniformly continuous real functions. Let $F:\mathbb R^2\to\mathbb R$ be defined by

$$F(x_1,x_2)=\begin{cases} \frac{f(x_2)-f(x_1)}{x_2-x_1} & \text{if }x_1\neq x_2 \\ f'(x_1) & \text{if }x_1=x_2. \end{cases}$$

Is there a simple proof (i.e. one that proceeds directly from the definitions of uniform continuity and limits) that $F$ is also uniformly continuous?

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Update: This has been proved by user251257 below, but I am still very interested in a simper proof that does not use the mean value theorem, if that is possible.

Answer 1

Let $\epsilon > 0$. Then, there exists a $\delta > 0$ such that for every $x, \tilde x\in \mathbb R$ with $|x-\tilde x| < \delta$ follows $|f(x) - f(\tilde x)| < \epsilon$ and $|f'(x) - f'(\tilde x)| < \epsilon$.

Now, let $x_0, x_1, x_2\in\mathbb R$ with $|x_i - x_0| < \delta$ for $i=1,2$ and $x_1 \ne x_2$. Then, for every $t\in [0, 1]$ follows $$ |(1-t)x_1 + tx_2 - x_0| \le (1-t)|x_1 - x_0| + t|x_2 - x_0| < \delta. $$ By mean value theorem there exists a $t\in[0,1]$ with \begin{align*} |F(x_1, x_2) - F(x_0, x_0)| &= \left| f'((1-t)x_1 + tx_2) - f'(x_0) \right| < \epsilon. \end{align*}

Do you have a guess for the other cases?