Negation of monotonicity 
Let $f:D \longrightarrow \mathbb{R}$ be a function where $D \subset \mathbb{R}$ which is not monotonic. Then there exists a 'cap' or a 'cup' in its graph, namely, there are $a,b,c \in D$ such that $a<b<c$ and  $f(a), f(c) < f(b) \vee f(b) < f(a), f(c)$.

Ok. This is easy. Easy enough so that we have frequently seen the phrase saying that '…so, evidently, there are three numbers $a<b<c$ so that either $f(a), f(c) < f(b)$ or $f(b) < f(a), f(c)$' at the stage where the function $f$ is not monotonic. However, this is rarely had been proved. I guess it is high time to search for a 'good' and/or 'short' and/or 'easy' proofs.
 A: Constant functions are monotonic, so suppose $f$ is not constant.  Thus there are $a,b$ with $a < b$ and $f(a) \ne f(b)$.  We may suppose $f(a) < f(b)$ (otherwise consider $-f$ instead of $f$).
Suppose there are no "cups" or "caps". I claim $f$ is non-decreasing.  


*

*If $x \in D$ with $x < a$, then for $(x,a,b)$ to not be a "cup" we need $f(x) \le f(a)$.  Moreover, if $x_1, x_2 \in D$ with $x_1 < x_2 < a$, since $\max(f(x_1), f(x_2)) \le f(a)$, for $(x_1, x_2, a)$ to not be a "cup" we need $f(x_1) \le f(x_2)$.  Thus
$f$ is nondecreasing on $D \cap (-\infty, a]$.

*A similar argument shows that $f$ is nondecreasing on $D \cap [b,\infty)$.

*If $x \in D$ with $a < x < b$, then for $(a, x, b)$ to be neither "cup" nor "cap" we need $f(a) \le f(x) \le f(b)$.  Moreover, if $x_1, x_2 \in D$ with $a < x_1 < x_2 < b$, for
$(a, x_1, x_2)$ not to be a "cap" we need $f(x_1) \le f(x_2)$.
Thus $f$ is nondecreasing on $D \cap [a,b]$.


Putting these together, $f$ is nondecreasing on all of $D$.
A: Let $x,y,x',y'\in D$ with $x<y$ and $f(x)<f(y)$ and $x'<y'$ and $f(x')>f(y').$
$I.$ If $y'\ge y.$
$I(i).$ If $f(y')<f(y)$ then $y'>y$ so let $a=x, b=y, c=y'.$
$I(ii).$ If $f(y')\ge f(y)$ then $f(x')>f(y)>f(x),$ so 
$\quad(ii-a).$ If $x'>y$ let $a=x, b=x', c=y'.$
$\quad (ii-b).$ If $x<x'<y$ let $a=x, b=x', c=y.$
$\quad (ii-c).$ If $x'<x$ let $a=x', b=x, c=y.$
Note: In $I(ii)$ the cases $x'=y\lor x'=x$ do not occur, as $f(x')>f(y)>f(x).$
$II.$ If $y'<y.$ In $I.$ replace $f$ with $-f,$ where $(-f)(z)=-f(z)$ for $z\in D,$ and replace $(x,y,x',y')$ with $(x',y',x,y).$ Then by $I$ the graph of $-f$ has a cap or a  cup so the graph of $f$ does too.  
A: Choose $s_1,s_2\in D$ with $s_1\lt s_2$ and $f(s_1)\lt f(s_2)$; choose $t_1,t_2\in D$ with $t_1\lt t_2$ and $f(t_1)\gt f(t_2)$; and let $E=\{s_1,s_2,t_1,t_2\}$. Let $x_L=\min E$, $x_R=\max E$, and $M=\max\{f(x):x\in E\}$.
Case 1. $f(x_L)=M$.
Let $a=x_L$, $b=s_1$, $c=s_2$. Then $a\lt b\lt c$ and $f(a)\gt f(b)\lt f(c)$; we have a cup.
Case 2. $f(x_R)=M$.
Let $a=t_1$, $b=t_2$, $c=x_R$. Then $a\lt b\lt c$ and $f(a)\gt f(b)\lt f(c)$; we have a cup.
Case 3. $f(x_L)\lt M$ and $f(x_R)\lt M$.
Let $a=x_L$, let $c=x_R$, and choose $b\in E$ with $f(b)=M$. Then $a\lt b\lt c$ and $f(a)\lt f(b)\gt f(c)$; we have a cap.
A: Be $f$ not monotonic.
Take two points $a<b$ with $f(a) < f(b)$ (those exist because otherwise the function would be monotonic decreasing), and two points $c<d$ with $f(c) > f(d)$ (if those don't exist, the function is monotonic increasing).
Now assume there were no three points among them that form either a cup nor a cap.
Consider the point $u\in\{a,b,c,d\}$ with the least function value. If it were in between two other points, we would have a cup, therefore it is either the smallest or the greatest of the four points. Let's without loss of generality assume that it is the smallest (reversing the order of points doesn't add or remove caps or cups).
Now consider the point $x\in\{a,b,c,d\}$ with the largest function value. The analogous argument applies here, so it has to occupy the other end.
Now consider the two remaining points. Let's call the smaller one $v$ and the greater one $w$ (so we have $u\le v\le w\le x$). Now clearly if $f(v)>f(w)$, then the points $u,v,w$ form a cap and the points $v,w,x$ form a cup. Since we assumed that neither cup nor cap exists, we have therefore $f(v)\le f(w)$.
But that means we have $u\le v\le w\le x$ and $f(u)\le f(v)\le f(w)\le f(x)$. In particular, there are no two points $p,q \in \{u,v,w,x\}$ such that $p<q$ and $f(p)>f(q)$. But by construction, $\{u,v,w,x\}=\{a,b,c,d\}$ (we just renamed those points), and by choice of points, $c<d$ and $f(c)>f(d)$. Therefore the assumption that no cups or caps exist among those four points cannot hold.
