Problems with a proof that every real sequence has a monotonic subsequence I have some problems with the visualization of this proof.
(I will present the problems I have with it in the end, with some intuitive thoughts related to them in italics.)
Theorem: Every real sequence has a monotonic subsequence.  
Proof (Thurston):
Take any $(x_m) \in R^\infty$ and define $S_m := \{x_m, x_{m+1},\dots \}$ for each $m \in \mathbb{N}$. If there is no maximum element in $S_1$, then it is easy to see that $(x_m)$ has a monotonic subsequence. (Let $x_{m_1} := x_1$, let $x_{m_2}$ be the first term in the sequence $(x_2,x_3,\dots)$ greater than $x_1$, let $x_{m_3}$ be the first term in the sequence $(x_{m_{2}+1}, x_{m_{2}+2}, \dots)$ greater than $x_{m_2}$ , and so on.) By the same logic, if, for any $m \in \mathbb{N}$, there is no maximum element in $S_m$, then we are done. Assume then max $S_m$ exists for each $m \in \mathbb{N}$. Now define the subsequence $(x_{m_k})$ recursively as follows:
$$ x_{m_1} := \text{max} S_1,\hspace{0.7cm} x_{m_2} := \text{max} S_{m_{1}+1}, \hspace{0.7cm}x_{m_3} := \text{max} S_{m_{2}+1}, \hspace{0.7cm} \dots $$
Clearly, $(x_{m_k} )$ is decreasing. QED
Problems 
1) Why do we have to build $S_m$ for any $m > 1$?
I mean, to me it looks enough to do the construction in brackets for $S_1$, also because $S_m$ should be - by construction - a subset of $S_1$, right?
Basically, I have a lot of problems with the sentence "By the same logic, if, for any $m \in \mathbb{N}$, there is no maximum element in $S_m$, then we are done.". To me, we are done way before.
2) What about the fact that $x_m = x_{m+1}$?
Shouldn't it be covered explicitly? And isn't it actually explicitly ruled out (i.e. the use of "greater" without mentioning equality) by the proof?
Indeed, nowhere it is assumed that we are talking about strictly monotonic subsequences (this looks really as an hidden assumption to me). 
3) Do we build up the $S_m$ set for any $m >1$ because we have to use it in the second part of the proof?  
Many thanks for any hint or feedback.
 A: 1) If you take the sequence $10,20,30,1 - \frac{1}{4}, 1 - \frac{1}{5}, \ldots 1 - \frac{1}{n}, \ldots$, $S_1$, $S_2$ and $S_3$ all have a maximum element, whereas $S_4$ does not have one. In other words, if $S_m$ has a maximum element, so does $S_1$; but the converse is not true.
2) I believe that the word "greater" here should read "greater than or equal to". Of course,  the sequence $1,1,1,1, \ldots $ has no strictly monotonic subsequence.
3) The $S_m$ are useful constructions for both parts of the proofs. Can you clarify the problems you had with it?
A: 1) You need to be able to keep picking members of the sequence that are further on than the ones you have already used in the subsequence you're constructing - hence needing to keep considering $S_m$ (but yes, they will all be subsets of $S_n$ for all $n < m$ - and you need this fact to know that the last sequence will actually be decreasing).
2) The theorem is not true for all sequences if you insist on a strictly monotonic subsequence - consider the sequence 1,1,1,1,... for example.
A: 1) $S_1$ may have a maximum. 
2) Constant sequences seem are considered monotonic (though they are not strictly monotonic).
3) If we find a single $S_m$ without maximal element we are done. As it is possible that no $S_m$ has a maximal element we automatically have them all available for the fail case.
