I need to solve this:
$$f:\mathbb R\to\mathbb R,\mathcal C^1\text{-function and } a,b\in\mathbb R \text{ such that } a\lt b.$$
1) Probe that there is $M\gt 0$ such that $\forall x\in[a,b]$ it's verified that $ \vert f'(x)\vert \le M$
2) Conclude that $\forall x, y\in[a,b]$it's verified that $\vert f(x)-f(y)\vert \le M\vert x-y\vert$
Apparently the hypothesis placed upon $f:\Bbb R \to \Bbb R$ is that $f \in \mathcal C^1(\Bbb R, \Bbb R)$; then given $a, b \in \Bbb R$ with $a < b$ we have $f \in \mathcal C^1([a, b], \Bbb R)$, which implies that $f'(x)$ exists and is continuous on $[a, b]$; but a continuous function such as $f'(x)$ on a compact interval such as $[a, b]$ is necessarily bounded on that interval; thus we have the existence of $M$ such that
$x \in [a, b] \Longrightarrow \vert f'(x) \vert \le M, \tag 1$
and item (1) is established.
As for item (2), it follows from simple integration; we have for $x, y \in [a, b]$ with $x \le y$
$\vert f(x) - f(y) \vert = \vert f(y) - f(x) \vert = \displaystyle \left \vert \int_x^y f'(s) \; ds \right \vert$ $\le \displaystyle \int_x^y \vert f'(s) \vert \; ds \le \int_x^y M \; ds = M(y - x) = M \vert y - x \vert = M \vert x - y \vert; \tag 2$
and, for $y < x$,
$\vert f(x) - f(y) \vert= \displaystyle \left \vert \int_y^x f'(s) \; ds \right \vert \le \int_y^x \vert f'(s) \vert \; ds \le M \vert x - y \vert, \tag 3$
by exactly the same reasoning followed in (2); combining these two inequalities yields, for all $x, y \in [a, b]$,
$\vert f(x) - f(y) \vert \le M \vert x - y \vert. \tag 4$