A function $f:\mathbb{R}\rightarrow\mathbb{R}$ is called monotone if $f$ satisfies one of the following conditions $$ x_1,x_2\in\mathbb{R}, x_1<x_2\quad\Longrightarrow\quad f(x_1)\leq f(x_2) \quad (1) $$ $$ x_1,x_2\in\mathbb{R}, x_1<x_2\quad\Longrightarrow\quad f(x_1)\geq f(x_2) \quad (2) $$ We consider a condition to combine both (1) and (2) as following $$ x_1,x_2,x_3\in\mathbb{R}, x_1<x_2<x_3\quad \Longrightarrow\quad \min\{f(x_1),f(x_3)\}\leq f(x_2)\leq\max\{f(x_1),f(x_3)\}\quad (3) $$ It means that $f$ is monotone if and only if (3) is satisfied. Below is our proof of this fact.
($\Rightarrow$) Suppose that $f$ is monotone. Then either (1) or (2) is satisfied. Clearly, (3) holds.
($\Leftarrow$) Suppose that (3) holds. We will prove either (1) or (2) holds and so $f$ is monotone.
Suppose that $x_1<x_2$ and $f(x_1)\leq f(x_2)$. Let $x,y,z\in\mathbb{R}$ be such that $x<x_1<y<x_2<z$. We will prove that $f(x)\leq f(x_1)\leq f(y)\leq f(x_2)\leq f(z)$. Observe that
(3) and $x_1<y<x_2, f(x_1)\leq f(x_2)$ imply $f(x_1)\leq f(y)\leq f(x_2)$
(3) and $x<x_1<y, f(x_1)\leq f(y)$ imply $f(x)\leq f(x_1)\leq f(y)$
(3) and $y<x_2<z, f(y)\leq f(x_2)$ imply $f(y)\leq f(x_2)\leq f(z)$
Hence (1) holds if there exist $x_1<x_2$ such that $f(x_1)\leq f(x_2)$. If there exist $x_1<x_2$ such that $f(x_1)\geq f(x_2)$ then (2) is proved similarly to (1).
Our aim is to find a shorter proof and an intuition of this fact.