A simple counter example to "The product of two monotone sequences is a monotone sequence" is the product of the monotone sequence $\{...,-3,-2,-1,0,1,2,3,...\}$ (which you can picture as a sequence of points in the $y$ axes of the the graph of $f(x) =x$) with itself. The product of these sequences is again a sequence viewed as a series of points in the $y$ axis of $f(x)=x^2$, and is increasing for $n>0$ but decreasing for $n<0$, as can be easily checked. So this shows that even when both sequences are increasing, their product need not be monotone. However, one can easily check that if the sequences are both increasing or both decreasing, and neither change sign, their product is monotone.
Let $c=f(a), d=f(b) $ and $g=f^{-1}$. You can directly prove that $g$ is continuous on $[c, d] $. Let $p\in(c, d) $ so that $g(p) \in(a, b) $. Thus there exists a positive number $\epsilon_0$ such that $(g(p) - \epsilon_0,g(p)+\epsilon_0)\subseteq (a, b) $ (in particular you can take $\epsilon _0=\min(g(p)-a,b-g(p))$).
Consider an arbitrary $\epsilon >0$ and let $\epsilon'=\min(\epsilon, \epsilon_0)$. Then we have $(g(p) - \epsilon ', g(p) +\epsilon') \subseteq (a, b) $. It follows that $r=f(g(p)-\epsilon'),s=f(g(p)+\epsilon ') $ both lie in $(c, d) $ and $r<p<s$. Let us write $\delta=\min(p-r,s-p) $ so that $(p-\delta, p+\delta) \subseteq (r, s) $ and therefore $$g((p-\delta, p+\delta)) \subseteq g((r, s)) \subseteq (g(r), g(s)) =(g(p) - \epsilon', g(p)+\epsilon') \subseteq (g(p) - \epsilon, g(p) +\epsilon) $$ And this proves that $g$ is continuous at $p$. The proof can be easily modified/adapted for the case when $p=c$ or $p=d$.
The proof uses the fact that both $f, g$ are strictly increasing on their domains and you should be able to figure out where this has been used in above proof.
Best Answer
Monotone in words means it can only increase or it can only decrease. And in that case for instance a strictly increasing sequence is monotone since it cannot decrease.