Let $(\mathbb{K},0,1,+,\times,\leq)$ be an ordered field. Let $\delta$ be its cofinality, i.e. the length of the smallest sequence of $\mathbb{K}$ that is cofinal with it. $\delta$ is a regular cardinal. My question is the following :
Given a transfinite sequence in $\mathbb{K}$, $(x_\alpha)_{\alpha<\delta}$, can I extract a subsequence of length $\delta$ that is non-increasing or non-decreasing ?
In $\mathbb{R}$ for instance, $\delta=\omega$ (in my purpose I do not make any difference between cardinals and the corresponding ordinals). My statement is the monotone subsequence lemma. I just want a generalization with bigger fields.
I thinks that the fact that $\delta$ is a regular cardinal is important and using that $\delta$ is the cofinality of $\mathbb{K}$ has an important role to play (otherwise I would be able to find an increasing sequence of $\mathbb{R}$ that has an uncountable length). In particular, every dense subset has cardinality at least $\delta$.
As an additional hypothesis, I can assume that every Cauchy-sequence of length $\delta$ is convergent. Nevertheless $\mathbb{K}$ may not have the least upper bound property.
Edit : You can assume that $\mathbb{K}$ is real-closed if you want. I would be happy with it.
This property is called $\delta$-Ramsey for $\mathbb{K}$ in the article Generalized Archimedean fields by Cowles and LaGrange. It is true for all fields of cofinality $\delta$ if $\delta$ is a weakly compact cardinal (or if $\delta=\omega$). If $\delta$ is not, then it fails in certain fields.
Here is an example for $\delta=\omega_1$; which generalizes to any non weakly compact cardinal. Given an injective sequence $r=(r_{\alpha})_{\alpha<\omega_1}$ of real numbers, there may not be a monotonous subsequence of $r$. This is because this would entail the existence of an injective sequence $(q_{\alpha})_{\alpha<\omega_1}$ in $\mathbb{Q}$. Consider the field $\mathbb{R}((X_{\alpha})_{\alpha<\omega_1})$ obtained by adding inductively indeterminates $X_{\alpha}, \alpha<\omega_1$ to $\mathbb{R}$ and imposing $\mathbb{R}((X_{\alpha})_{\alpha<\beta})<X_{\beta}$ for all $\beta<\omega_1$. This has cofinality $\omega_1$ but the sequence $(r_{\alpha})_{\alpha<\omega_1}$ still has no monotonous subsequence in it. This is the same if you take its closure under limis of Cauchy $\omega_1$-sequences.
On the other hand, the field $\mathbb{Q}((X_{\alpha})_{\alpha<\omega_1})$ obtained similarly does have this $\omega_1$-Ramsey property. Cowles and LaGrange (and Sikorski before them) use this property along with the convergence of $\omega_1$-sequences and another property to obtain fields $F$ in which any $\operatorname{cof}(F)$-sequence has a convergent subsequence.
Regarding your quesiton about $\mathbf{No}(\delta)$.
$\mathbf{No}(\delta)$ is not Ramsey in general. This has cofinality $\delta$ (the cofinality of a linear order is the unique regular ordinal which embeds cofinally into it). Assume that there is a cardinal $\kappa <\delta$ with $2^{\kappa}\geq\delta$. This is not too wild an assumption. I choose $\kappa$ minimal, so for $\alpha < \kappa$, we have $2^{\alpha}<\delta$. In particular $\mathbf{No}(\kappa)$ has cardinal $<\delta$ by regularity of $\delta$. Its dense extension $D_{\kappa}$ of numbers in $\mathbf{No}(\kappa+1)$, in whose sign sequence the pattern $(-+)$ or $(+-)$ occurs cofinaly, has cardinal $2^{\kappa}\geq \delta$. Pick an injective sequence indexed by $\delta$ in $D_{\kappa}$. This does not have a monotonous subsequence, otherwise $\delta$ would order-embed in $\mathbf{No}(\kappa)$ by density, which cannot be by the previous cardinality argument. So $D_{\alpha} \subset\mathbf{No}(\delta)$, is a counterexample.
You can see that this is a generalization of the argument with real sequences indexed by $\omega_1$. There may be a simpler argument using a clever sequence of Hahn series, to deal with your other question, but I can't find one for now.