Every sequence in $\mathbb{R}$ has a monotonic subsequence I have trouble with this kind of infinite construction in topology. Can someone check my proof is sound?
Let $s$ be a sequence in $\mathbb{R}$. Then $s$ has a monotonic subsequence.
There are two cases. If $s$ is unbounded, it is easy. Without loss of generality, suppose $s$ is unbounded above. Choose a point $p_0$. For every $p_i$, there is some larger point $p_{i+1}$.
If $s$ is bounded, let $I$ be a closed interval containing all points in $s$. Then $I$ is compact and $s$ has a convergent subsequence $s^\prime$ which converges to $L$.
This is where I am not entirely sure of myself. 
Let $\varepsilon_0$ be any positive number. Choose a point $p_0$ within a radius of $\epsilon_0$ of $L$. Then take $\varepsilon_1$ to be a number smaller than $\varepsilon_0$ and choose a point $p1$ within $\varepsilon_1$. Repeat this to generate a convergent sequence $p_0, p_1, p_2, \dots$. 
Separate this sequence into $p_{i_0}, p_{i_1}, p_{i_2}, \dots$ of points less than $L$ and $p_{j_0}, p_{j_1}, p_{j_2}, \dots$ of points greater than $L$. It might be possible for one of these two sequences to be finite, but since $p_0, p_1, p_2, \dots$ was infinite, one of these two must also be infinite. Choose one which is infinite. It will be monotonic (either increasing or decreasing, depending on which was chosen).
 A: Let $(x_n)$ be a sequence in some linearly ordered set $(X,<)$. 
Call $n \in \mathbb{N}$ a peak of the sequence if for all $m > n$ we have $x_n > x_m$, i.e. $x_n$ is larger than all subsequent terms of the sequence.
Case 1: There are infinitely many peaks: suppose $n_1 < n_2 < n_3 < \ldots$ are infinitely many peaks of the sequence. Then by definition of a peak, $x_{n_1}, x_{n_2}, \ldots$ is a monotonically decreasing subsequence of $(x_n)$.
Case 2: There are only finitely many peaks. So there is some index $N$ that is the last peak. Then define $n_1 = N+1$. Then as $x_{n_1}$ is not a peak, there is some $n_2 > n_1$ such that $x_{n_2} \ge x_{n_1}$. Using recursion, we pick $n_{k+1} > n_k$ for all $k$, such that $x_{n_{k+1}} \ge x_{n_k}$, which is possible as $x_{n_k}$ is not a peak. This defines a increasing subsequence of $(x_n)$.
Alternatively, we apply the infinite Ramsey theorem using the two colours on $[\mathbb{N}]^2$ where $\{n,m\}$ has colour 0 iff $x_{\min(n,m)} \le x_{\max(n,m)}$ and 1 otherwise. A homogeneous subset of $\mathbb{N}$ defines a monotonic sequence.
This works in all ordered spaces, and the topology of $\mathbb{R}$ has nothing to do with it. In fact, this is often a lemma in the proof of the Bolzano-Weierstrass theorem that every bounded sequence of reals has a convergent subsequence (so using this very theorem 
in the proof of the simpler result seems circular, or overkill, or both).
A: As written, your choice of $p_0$ might b a lot closer to $L$ than $\varepsilon_1$ so you can'tknow that $p_1$ is larger than $p_0$ (if we assume - without loss of generality- that both are smaller). - It should be easy to fix that.
A: This will not work: Consider a sequence $p_0, p_1, …$ made by zipping $1/n$ and $1/2^n$ – still converging to zero (and only from above), but not monotonically.
You can fix this by working with rings instead of balls.
Take a limit point $L$ of the sequence $a_1, a_2, …$. You may assume that it has no constant subsequences. So either you will find an infinitiude of its points in every neighbourhood of $L$ and above $L$ or you will find an infinitude of its points in every neighbourhood of $L$ and below $L$. Without loss of generality, assume the latter. Define
$$R_n(L) = \{x ∈ ℝ;~1/2^{n+1} ≤ d(L,x) < 1/2^n\}.$$
Set $ν = 1$, run through $n ∈ ℕ$ and whenever there is some $a_k$ (with $k > k_{ν-1}$) in $R_n$ and below $L$, pick the first one and remember its index $k$ as $k_ν$, increase $ν$ by $1$ and move on to the next $n$. As there are infinitely many $a_k$ in each neighbourhood of $L$ and below $L$ you will have to remember infinitely many $k_ν$, so $(a_{k_ν})$ will be a monotically increasing subsequence of $(a_k)$.
A: Suppose wlog that the points $p_{i_0}, p_{i_1}, ...$ strictly less than $L$ are infinitely many and converges to $L$. Then you can build up this sequence. $a_0 = p_{i_0}$, $a_n \in (a_{n-1}, L)$ (of course taking $a_k$ in increasing order with respect to the indexes of $p_i$ sequence). This new sequence should work.
A: You must choose $\varepsilon_1$ to be smaller than $|p_0-L|$ as well as $\varepsilon_0$. And you must choose $p_i$ to be not equal to $L$. (If this last condition is impossible to satisfy, then there are an infinite number of terms equal to $L$, which themselves form a monotonic subsequence.)
