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 $\sum a_n t^n$.
I'm going to consider three questions:
What does the existence of the real numbers is needed?
Any reasoning which starts with the sentences above is nonsense because such a number and such a triangle don't exist. Analogously, if $\mathbb{R}$ (defined as an complete ordered field) did not exist (i.e. if there were no structure satisfying the axioms of complete ordered field) then the analysis course would be nonsense. This is why the proof of the existence is needed, to show that we are indeed doing something instead of nothing.
What does existence of the real numbers mean?
Real Analysis can be viewed as an axiomatic theory. In this context, the proof of the existence of $\mathbb{R}$ (i.e. the construction of a complete ordered field) means that there is a model for the axioms of a complete ordered field and thus the theory (i.e. the real analysis) is consistent.
Why would we need the completeness axiom?
Without the completeness axiom we cannot do analysis. Take, for example, $\mathbb Q$. It satisfies all axioms that defines a complete ordered field except the completeness axiom and nevertheless is inadequate for analysis.
Best Answer
The answer to your question depends critically on what you mean by a "complete ordered field" $(F,<)$. Here are two rival definitions:
[added: sequentially] Cauchy complete: every Cauchy sequence in $F$ converges.
Dedekind complete: every nonempty subset $S \subset F$ which is bounded above has a least upper bound.
(There are in fact many other axioms equivalent to 2): that every bounded monotone sequence converges, that $F$ is connected in the order topology, the principle of ordered induction holds, and so forth.)
It turns out that there is a unique Dedekind complete ordered field up to (unique!) isomorphism, namely the real numbers $\mathbb{R}$. Famously $\mathbb{R}$ is also Cauchy complete -- or, if you like, Dedekind complete ordered fields satisfy the Bolzano-Weierstass theorem, which is enough to make Cauchy sequences converge -- so that Dedekind completeness implies Cauchy completeness.
The converse is true with an additional hypothesis: an Archimedean Cauchy-complete field is Dedekind complete. I show this in $\S 12.7$ of these notes using somewhat more sophisticated methods (namely Cauchy nets). For a more elementary proof, see e.g. Theorem 3.11 of this nice undergraduate thesis.
On the other hand, just as one can take the "Cauchy" completion of any metric space (or normed field) and get a complete metric space (or complete normed field), one can take the Cauchy completion of a non-Archimedean ordered field and get an ordered field which is Cauchy complete but not Dedekind complete. The easiest example of such a field is probably the rational function field $\mathbb{R}(t)$ with the unique ordering that makes $t$ positive and infinitely large.
For some reason these subtleties seem to be hard to find in standard analysis texts. I myself didn't learn about them until rather recently (so, several years after my PhD). I actually wrote up some of this material as supplemental notes for a sophomore-junior level course I am currently teaching on sequences and series...but I have not as yet been able to make myself inflict these notes on my students. I talked about ordered fields in several lectures and it seemed to be one level of abstraction beyond what they could even meaningfully grapple with (so it started to seem a bit pointless).