The standard counterexample is the field $F$ of finitely-tailed Laurent series over $\mathbb{R}$: series of the form $\sum a_n t^n$ where $a_n \in \mathbb{R}$, exponents $n \in \mathbb{Z}$, but with only finitely many negative exponents (i.e., $a_n = 0$ for all but finitely many negative integers $n$.)
Addition and multiplication are defined just like for power series; you should verify that the "finitely many negative exponents" condition is essential to ensuring that multiplication of finitely-tailed Laurent series is well-defined. Check that non-zero elements of $F$ have multiplicative inverses: a series that starts with $a_n t^n$ has an inverse that starts with $a_n^{-1} t^{-n}$. (Here, and below, when I say that a nonzero element of $F$ starts with $a_n t^n$, I mean that $n$ is the lowest integer for which $a_n \ne 0$.)
Ordering is defined as follows: An element is positive if its starting term has positive coefficient. It's easy to check: (1) for every nonzero $x \in F$, exactly one of $x$ and $-x$ is positive; (2) the sum and product of 2 positive elements is positive.
So $F$ is an ordered field. Note that $F$ is not Archimedean: $t^{-1}$ is bigger than every integer. The field $F$ also does not satisfy the least upper bound property: $t^{-1}$ is an upper bound for the integers, but there is no least upper bound (check that if $x$ is such an upper bound, then $x/2$ is a smaller one.)
To discuss Cauchy completeness, we have to put a metric on $F$, as follows. Pick any positive real number $q>1$. If a nonzero element $x \in F$ starts with $a_n t^n$, then define $|x| = q^{-n}$ (and of course define $|0|=0$), and then define the distance between $x$ and $y$ to be $|x-y|$. Check that this is a metric space. (Caution: This doesn't extend the existing metric on $\mathbb{R}$; all real numbers have the same size under this metric.) In fact, it makes $F$ into a valued field; we have $|xy| = |x||y|$.
This is a "non-Archimedean" metric because it satisfies the strong triangle inequality $d(x,z) \le \max(d(x,y), d(y,z))$. For a non-Archimedean metric, a sequence is Cauchy iff $|x_n - x_{n+1}| \to 0$ as $n \to \infty$.
To prove that $F$ is complete under this metric, note that for a Cauchy sequence in $F$, the coefficients for a fixed exponent $t^n$ must be eventually constant; otherwise the difference between consecutive terms could never be smaller than $q^{-n}$. Call this constant $a_n$; then the Cauchy sequence converges to $\sum a_n t^n$.
In an arbitrary ordered field one has the notion of Dedekind completeness -- every nonempty bounded above subset has a least upper bound -- and also the notion of sequential completeness -- every Cauchy sequence converges.
The main theorem here is as follows:
For an ordered field $F$, the following are equivalent:
(i) $F$ is Dedekind complete.
(ii) $F$ is sequentially complete and Archimedean.
Since the Archimedean hypothesis often goes almost without saying in calculus / analysis courses, many otherwise learned people are unaware that there are non-Archimedean sequentially complete fields. In fact there are rather a lot of them, and they can differ quite a lot in their behavior: e.g. some of them are first countable in the induced (order) topology, and some of them are not. (In particular there are some ordered fields in which a sequence is Cauchy if and only if it is eventually constant! This is a case where one should consider Cauchy nets if one is serious about exploring the topology...)
These issues are treated in $\S 1.7$ and $1.8$ of these notes.
Best Answer
First we will establish some intermediate results:
Theorem 1: Every Cauchy sequence is bounded.
Proof: Take $\epsilon=1$, then there is $N\in\mathbb{N}$ such that
$$m, n \geq N\quad\implies |x_m-x_n|<1$$
So that $$|x_m|<|x_N|+1\quad\quad\forall\;m\geq N$$ therefore $$|x_n|\leq\max\lbrace |x_1|,\ldots,|x_N|,|x_N|+1\rbrace\quad\quad\forall\;n\in \mathbb{N}$$
Theorem 2: Every sequence has a monotone subsequence.
Proof: Lemma in Bolzano-Weierstrass
Theorem 3: Let $F$ be an ordered field with the LUB property, then every bounded and monotone sequence in $F$ is convergent.
Proof: WLOG lets assume the sequence is increasing. As it is bounded it has a supremum: $s=\sup_{n\in\mathbb{N}}\lbrace x_n\rbrace$ We claim that $s$ is the limit of the sequence.
Let $\epsilon>0$ then there is $N\in\mathbb{N}$ such that
$$s-\epsilon<x_N\leq x_{N+1}\leq \cdots \leq s$$
so that for any $n\geq N$ we have $$|x_n-s|<\epsilon$$
Which proves that $s$ is the limit.
To simplify discussion, we assume that the order field is $\mathbb{R}$.
For the first implication $(1)\implies (2)$:
To establish the Archimedean Property first note that the set of integers is not bounded above, if it were it would have a supremum. Let's call it $z$; then there would be an integer $n$ such that $z-1<n$ but then $z<n+1\in \mathbb{Z}$ This contradicts the fact that $z$ was the supremum.
Archimedean Property: For every real $x$ there is an integer $n$ such that $n>x$
Proof: If this is not the case, integers would be bounded by some $x$.
Theorems $(1)$, $(2)$ and $(3)$ above prove Cauchy completeness.
For the reverse implication $(2)\implies (1)$:
Theorem 4: Suppose $F$ has the Archimedean property, then every monotone bounded sequence is Cauchy.
Proof: WLOG take the sequence to be increasing. If $\lbrace x_n\rbrace_{n}$ is not Cauchy then there exists $\epsilon>0$ so that for every $N\in\mathbb{N}$
$$n>m\geq N\quad\implies x_n-x_m\geq \epsilon$$
We are going to extract a subsequence such that it is not bounded.
For $N=1$ choose ${n_1},\ {n_2}$ such that $x_{n_2}-x_{n_1}>\epsilon$ Now take $N^{\prime}>n_2$ and choose ${n_3},\ {n_4}$ such that $x_{n_4}-x_{n_3}>\epsilon$. Continue in this way to construct a subsequence. Note that there are infintely many differences greater than $\epsilon$, so by the Archimedean Principle the subsequence diverges. This contradicts the fact that the sequence was bounded.
Observe that if $F$ is also Cauchy complete then every monotone bounded sequence is convergent.
We have established:
Cauchy completeness+Archimedean Property $\implies$ Convergence of every monotone and bounded sequence
The answer in this post proves:
Convergence of every monotone and bounded sequence $\implies$ LUB Property