Understanding the proof that $\cosh$ is strictly increasing on $[0,\infty)$ I need help understanding this proof:
Prove that $\cosh$ is strictly increasing on $[0,\infty)$.

From $(\cosh x)^2=1+(\sinh x)^2\geq 1$ it follows that $R(\cosh)\subseteq [1,\infty)$. From that for $x,y>0$ we have: $\cosh (x+y)=\cosh x \cdot \cosh y+\sinh x\cdot \sinh y> \cosh x\cdot \cosh y> \cosh x$ so $\cosh$ is strictly increasing on $[0,\infty)$. 

$1)$ Why is $R(\cosh)\subseteq[1,\infty)$? If $(\cosh x)^2\ge 1$ doesn't that mean $R(\cosh x)=(-\infty, -1]\cup [1,\infty)$? Then I would say $[1,\infty)\subseteq R(\cosh x)$ and not the other way around? 
$2)$
I understand that $\cosh (x+y)=\cosh x \cdot \cosh y+\sinh x\cdot \sinh y> \cosh x\cdot \cosh y> \cosh x$ but I don't see how that proves $\cosh x$ is strictly increasing on $[0, \infty)$... If we want to prove that a function is strictly increasing, then I thought we should take $a, b \in D(f)$ and prove that if $a<b \Rightarrow f(a)<f(b)$. Can someone explain the idea behind this proof? Thank you!
 A: In $1)$, the knowledge that $\cosh x \geqslant 0$ for all $x\in \mathbb{R}$ (or only for $x \geqslant 0$) is used. With that knowledge, from $(\cosh x)^2 \geqslant 1$ one can infer $\cos x \geqslant 1$, i.e. $R(\cosh) \subseteq [1,\infty)$. Indeed $R(\cosh) = [1,\infty)$, but showing equality requires a further step. For example with $f(x) = \sqrt{1 + \lfloor x^2\rfloor}$ we also can deduce $R(f) \subseteq [1,\infty)$ from the fact that $f(x) \geqslant 0$ for all $x$ and $f(x)^2 = 1 + \lfloor x^2\rfloor \geqslant 1$, but here $R(f) \subsetneq [1,\infty)$.
Regarding point $2)$, for $0 \leqslant a < b$, take $x = a$ and $y = b-a$, then the argument yields
$$\cosh b = \cosh (x+y) > \cosh x = \cosh a.$$
In the argument, the step $x > 0 \implies \cosh x > 1$ is however missing.
A: Here is an alternative proof: Set $a=\exp x_1$ and $b=\exp x_2$ where $x_1,x_2 \in [0,\infty)$ and $x_1<x_2$. Then we have $1\leq a <b$ since $\exp$ is increasing and we have
$$ 0< \frac{1}{a}- \frac{1}{b} = \frac{b-a}{ab} < \frac{b-a}{1}=b-a.$$
The inequality is equivalent to
$$a + \frac{1}{a} < b + \frac{1}{b}.$$
Therefore we have
$$\cosh x_1 < \cosh x_2$$
since $\cosh x= \frac{1}{2}( \exp x+ \exp(-x))$.
