I'm trying to solve the following exercise.
Let $K$ be an ordered field and $(x_n)_{n\in \mathbb{N}}$ be an Cauchy sequence in $K$. Show that there exists an increasing Cauchy sequence $(y_n)_{n\in \mathbb{N}}$ and a decreasing Cauchy sequence $(z_n)_{n\in \mathbb{N}}$ with $$x_n=y_n+z_n.$$
If the field were complete, it would be easier. But I'm not sure how to construct $(y_n)_{n\in \mathbb{N}}$ and $(z_n)_{n\in \mathbb{N}}$ without an existing limit.
On
Given the ability to refer to the limit of $(x_n)_{n\in\mathbb{N}}$ in a complete metric space, I would imagine that your idea for a proof would refer to the decreasing sequence of points above the limit (call this set $A'$), and the increasing set of points below the limit (call this set $B'$). Even if the space $K$ is not complete, we can still argue that $A'$ and $B'$ are subsets of the space! This is deeply related to the Dedekind cut method of completing a total linear ordering, and I encourage you to think about the analogy between this construction and ours! In fact, the notation $A'$ and $B'$ is chosen to build this analogy with the notation chosen in the wikipedia page for Dedekind cuts, but reflect that the construction is not exactly the same.
Now we can go about constructing $A'$ and $B'$. First, let's make the choice that if the limit of the sequence exists as an element of the sequence, we can include it in the set $B'$. We can then describe the set $A'$ as the subset of $(x_n)_{n\in\mathbb{N}}$ including all points with only finitely many points from $(x_n)_{n\in\mathbb{N}}$ below them and $B'$ as the subset of $(x_n)_{n\in\mathbb{N}}$ including all points with countably infinitely many points from $(x_n)_{n\in\mathbb{N}}$ below them. If you're unfamiliar with formal set theory, this works as a good informal construction of $A'$ and $B'$. For the set theorists, this also works as a formal description: by the separation axiom of set theory, we can write a first-order sentence to describe both of these properties and cut the subsets $A'$ and $B'$ out of the set of points $(x_n)_{n\in\mathbb{N}}$.
The rest of the proof should then run as if we had a limit point to refer to!
The statement seems to be wrong. Let's consider $K= \mathbb{R}$ (which is an ordered field) and $x_n=(-1)^n/n$. The sequence $(x_n)_n$ converges (to $0$) and thus is a Cauchy sequence. Assume that the exists a monotone increasing sequence $(y_n)_n\subseteq \mathbb{R}$ and a monotone decreasing sequence $(z_n)_n\subseteq \mathbb{R}$ such that $x_n= y_n + z_n$. Then we get
$$ x_{n+1} -x_n = (y_{n+1}-y_n) +(z_{n+1}-z_n). $$
As $(z_n)_n$ is decreasing, we get that $z_{n+1}-z_n\leq 0$ and hence, we have
$$ y_{n+1}-y_n \geq x_{n+1}-x_n. $$
Again, as $y_{k+1}-y_k \geq 0$ we get
\begin{align*} y_{2n} &= y_1+\sum_{k=1}^{2n-1} (y_{k+1}-y_{k}) \geq y_1+ \sum_{j=1}^{n} (x_{2j}-x_{2j-1}) \\ &= y_1+ \sum_{j=1}^n \left( \frac{1}{2j} + \frac{1}{2j-1} \right) \geq y_1 + \frac{1}{2} \sum_{j=1}^n \frac{1}{j}. \end{align*}
Hence, we get that $\lim_{n\rightarrow \infty} y_{2n} = \infty$ and $(y_n)_n$ cannot be a Cauchy sequence in $\mathbb{R}$.