Every partial order has well-ordered chains, for example, singletons, or the empty chain.
Presumably you're asking about maximal well-ordered chains. And the answer to that is indeed positive.
It is easy to prove Zorn's lemma from this assumption: Assume that every partial order has maximal well-ordered chains, and let $(P,\leq)$ be a partial order in which every chain has an upper bound. If $C$ is a maximal well-ordered chain, let $x$ be an upper bound for it; if $x\notin C$, show that $C\cup\{x\}$ is still a well-ordered chain, so this is impossible, and therefore $x\in C$. From this follows that $x$ must be maximal, otherwise there would be an upper bound of $C$ which is not in $C$, which we established is impossible.
If what you're suggesting is "Every partial order whose chains are all well-ordered is well-founded", this is not enough to prove the axiom of choice.
To see that, first we need to prove another auxiliary choice principle: Dependent Choice, which is weaker than the axiom of choice. Dependent Choice can be stated as the principle: If $(P,\leq)$ is a partial order such that there is no infinite decreasing sequence, then $P$ is well-founded.
Now, assume Dependent Choice holds. If a partial order is not well-founded, then it has a decreasing sequence, which is a chain that is by definition not well-ordered, and therefore not all chains are well-ordered.
In the other direction, suppose that every partial order whose chains are well-ordered is well-founded, and let $(P,\leq)$ be a partial order without any decreasing sequences. Then consider $(Q,\prec)$ where $Q$ is the set of all decreasing sequences in $P$, ordered by reverse end-extensions (i.e., $\vec b\prec\vec a$ if $\vec a$ is an initial segment of $\vec b$).
It is not hard to verify that every chain in $Q$ is well-ordered, and therefore $Q$ is well-founded. Now suppose that $A$ is any subset of $P$, look at all the decreasing sequences in $A$, this is a subset of $Q$, so it has a minimal element, which is a decreasing sequence in $A$ that cannot be extended any further inside $A$, but those are finite sequences, and therefore this minimal sequence has an element which is minimal in $A$.
Best Answer
Assuming the axiom of choice holds, it is possible to well-order every set. In particular the real numbers.
Fix a choice function on $P(\mathbb R)\setminus\{\varnothing\}$, let us denote it by $f$. We now define by transfinite induction an injection from $\mathbb R$ into the ordinals:
We immediately have that $r_\alpha\neq r_\beta$ for $\alpha\neq\beta$; this has to terminate because $\mathbb R$ is a set, and the induction cannot go through the entire class of ordinals; and the induction covers all the real numbers, because we can keep on choosing.
One can appeal to equivalents of the axiom of choice to show existence:
Using Zorn's lemma, let $(P,\leq)$ be the collection of well-orders of subsets of the real numbers, ordered by extensions. Suppose we have a chain of such well-orders, their union is an enumerated union of well-ordered sets and therefore can be well-ordered (without assuming the axiom of choice holds in any form).
By Zorn's lemma we have a maximal element, and by its maximality it is obvious that we have well-ordered the entire real numbers.
Using the trichotomy principle (every two cardinals can be well-ordered) we can compare $\mathbb R$ with its Hartogs number $\kappa$ (an ordinal which cannot be injected into $\mathbb R$), it has to be that $\mathbb R$ injects into $\kappa$, and therefore inherits a well-order by such injection.
The list goes on. The simplest would be to use "The power set of a well-ordered set is well-ordered". As $\mathbb N$ is well-ordered, it follows that $\mathbb R$ can be well-ordered.
However no other proof that I know of has any sense of constructibility as the use of a choice function on the power set of $\mathbb R$ and transfinite induction.