I can prove no descending chain implies well ordering. Take any subset $S$, assume it has no least element, then we can construct a function $f: S \to S$ such that $f(x) < x$, then use recursion theorem to prove there is a descending chain.
However the construction of $f$ requires axiom of choice. Is this needed? Or is there proof without AC?
To remove this from the unanswered queue:
Bof's comment is exactly right. In Cohen's original model of $\mathsf{ZF+\neg AC}$, we have a set of reals $X$ (namely the set of Cohen reals added to the model) which is not well-ordered but has no infinite descending sequence.