If $f\colon (0,1)\to(0,1)$ is surjective, is there necessarily an ordered (incr. or decr.) subsequence $\{x_i\}$ in the domain st $f(x_i)\to1$? Let $f\colon (0,1)\to(0,1)$; $f$ is surjective.
The claim is that, for $f$, there is either a strictly increasing or decreasing subsequence of $x_i$ in the domain for which $f(x_i)$ approaches $1$. 
I made this question up, so I am not even sure whether this claim is true. If it is true, I'd like to prove it.  
In the process of my work, I've become quite confused about how two types of proof, direct and proof by contradiction, relate to this problem.
Here is my approach:
I first choose an $x_1$ in the domain $(0,1)$, say $x_1=1/2$, and then head toward increasing values of $x$ in the domain and select a subsequent $x_i$ each time I encounter an $x$ st $f(x)>f(x_{i-1})$. If at some point, $f(x)<f(x_{i-1})$ for all $x>x_{i-1}$, I could switch directions, reset $x_1=x_{i-1}$ and begin all over again, knowing that as $f$ is surjective there is certainly a value of $x$ st $f(x)>f(x_1)$ in this new direction. I could continue this process, changing direction and resetting my $x_1$ when needed, for eternity. 
As far as a direct proof is concerned, my take is that a valid proof would entail deriving a complete subsequence for some increasing or decreasing ordering of $x$ values in the domain st $f(x_i)\to1$. I do not believe my work accomplishes this, however, my question is whether it has any merit as a proof by contradiction, specifically, that the claim is proven because each subsequence that reaches a "dead end" can be transformed (at least temporarily) into a new ordered subsequence by resetting my $x_1$ and heading in the other direction. I am hoping somebody can set me straight on whether this too fails, and clear up the general confusion I have with this problem. 
Thanks!
 A: It could be necessary to switch again and again and always afer finitely many steps. But as the function values always increase, you will never exceed a previous turning point on your way back. This ensures that the left turning points form an increasing sequence (with increasing function valkues) and the right turning points form a decreasing sequence (also with increasing function valkues). So at least we conclude that there exsts an infinite ordered sequence $x_i$ such tat $f(x_i)$ is strictly increasing. However, even then it may be the case that the function values stay way below $1$.
So a little extra care is needed.
First of all, there exists some sequence $x_i$ such that $f(x_i)\to 1$. This sequence has at least one limit point $x\in[0,1]$. By restricting to a subsequence if necessary we may assume wlog. that $x_i\to x$. There are infinitely many $i$ with $x_i>x$ or infinitely many $i$ with $x_i<x$. In the first case there is a strictly decreasing subsequence, in the latter a strictly increasing subsequence. And we still have the the function values conveger (monotonically!) to $1$.
