This question was actually asked by John Stillwell in a comment to an answer to this question. I thought I would advertise it as a separate question since no one has yet answered and I am also curious about it.
Question: Is the existence of a non-principal ultra-filter on $\omega$ a weaker assumption than the existence of a well-ordering of $\mathbb{R}$?
It is consistent that there exists a non-principal ultrafilter over $\omega$ while $\mathbb{R}$ is not well-ordered. To see this, suppose that the partition relation $\omega \to (\omega)^{\omega}$ holds in $L(\mathbb{R})$. Then forcing with $\mathbb{P}= [\omega]^{\omega}$ adjoins a selective ultrafilter $\mathcal{U}$ over $\omega$ and $\mathcal{P}(\omega)$ cannot be well-ordered in $L(\mathbb{R})[\mathcal{U}]$. (See Eisworth's paper: Selective ultrafilters and $\omega \to (\omega)^{\omega}$.) Thus $L(\mathbb{R})[\mathcal{U}]$ is a model of $ZF$ which contains the nonprincipal ultrafilter $\mathcal{U}$ and yet $\mathbb{R}$ cannot be well-ordered in $L(\mathbb{R})[\mathcal{U}]$.
An update: it is perhaps also interesting to note that in $L(\mathbb{R})[\mathcal{U}]$, the ultraproduct $\prod_{\mathcal{U}} \bar{\mathbb{F}}_{p}$ of the algebraic closures of the fields of prime order $p$ is not isomorphic to $\mathbb{C}$.
Historically it was proved that the ultrafilter lemma is independent from the axiom of choice by showing that there is a model in which there is an infinite Dedekind-finite set of real numbers, but every filter can be extended to an ultrafilter. Where an infinite Dedekind-finite set is an infinite set which does not have a countably infinite subset.
The existence of infinite Dedekind-finite sets negates not only the axiom of choice, but also the [much] weaker axiom of countable choice. These sets cannot be well-ordered, and since the real numbers have such subset they cannot be well-ordered themselves in such model.
The proof was given by Halpern and Levy in 1964.
