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
You can do this as follows.
Let $A\subset\mathbb R$ be nonempty and bounded above. Choose $a_0\in A$ and an upper bound $b_0$ for $A$. Then, look at the middle point $m$ of the interval $[a_0,b_0]$. If $m$ is an upper bound for $A$, put $b_1:=m$ and $a_1:=a_0$. Otherwise, put $b_1:=b_0$ and choose a point $a_1\in A$ such that $a_1>m$. In either case, you have $a_0\leq a_1\leq b_1\leq b_0$, and $b_1-a_1\leq \frac12 \,( b_0-a_0)$. Moreover, $a_1\in A$ and $b_1$ is an upper bound for $A$.
Repeating this procedure, you can construct by induction a non-decreasing sequence $(a_n)\subset A$ and a non-increasing sequence $(b_n)$ of upper bounds for $A$, with $a_n\leq b_n$ for all $n$ and $b_{n+1}-a_{n+1}\leq \frac12 (b_n-a_n)$.
By the Archimedean property (which you do need as pointed out by Daniel), the diameter of the interval $[a_n,b_n]$ goes to $0$. It follows that both sequences $(a_n)$ and $(b_n)$ are Cauchy. So they are both convergent, to the same limit since $b_n-a_n\to 0$. If you call this limit $l$, then $l$ is an upper bound for $A$ because the set of all upper bounds for $A$ is closed and $b_n\to l$, and no upper bound for $A$ can be smaller than $l$ because $l$ is in the closure of $A$.