If a function is continuous and one-to-one then it's strictly monotonic 
Let $f:(A,B)\to \mathbb{R}^1$ be continuous and one-to-one.
  Prove that $f$ is strictly monotonic on $(A,B)$.

Proof: 
Suppose $f$ is not strictly monotonic on $(A,B)$. Then there exist $a<b<c$ such that $f(a)\geqslant f(b)$ and $f(c)\geqslant f(b)$. 
Suppose that $f(a)\geqslant f(c)$. Since $[a,b]$ is connected and $f$ is continuous then $f([a,b])$ is connected. 
It's easy to prove that $[f(b),f(a)]\subset f([a,b])$.
If $f(b)=f(a)$, then $f$ is not one-to-one and we have contradiction.
If $f(b)<f(a)$, then $f(c)\in f([a,b])$ and by the intermediate value theorem $\exists c'\in [a,b]$ such that $f(c')=f(c)$ and $f$ is not one-to-one.
Sorry if this topic repeated but I would like to know is my proof correct?
Thanks in advance.
 A: Like BrianO said, your reduction to studying how $f$ behaves with three points is a little quick. However, the reduction is elementary in the sense that it only uses ordered field axioms and not the LUB or other analytic theorems.
Let me just give the details of the reduction:

A map is monotone iff:
$(i)$: it is monotone on every set of four points or less. 
This is just figuring out the negation of being monotone: if $f$ is not monotone then there are points $a \leq b$, $c \leq d$ in its domain such that $f(a) > f(b)$ (non increasing) and $f(c) < f(d)$ (non decreasing) and $f$ is not monotone on $\{a;b;c;d\}$.
$(ii)$: it is monotone on every set of three points or less. 
Indeed, assume $f$ satisfies $(ii)$. Let $a < b < c < d$ be four points with for instance $f(a) \leq f(b) \leq f(c)$.
If $f(a) = f(b) = f(c)$ then $f$ is monotone on $\{a;b;c;d\}$: increasing if $f(d) \geq f(a)$ and decreasing otherwise. Else, if for instance $f(a) < f(c)$, then applying $(ii)$ to $\{a;c;d\}$ yields $f(a) < f(c) \leq f(d)$ so $f$ is increasing on $\{a;b;c;d\}$. The three other cases work the same.
