Can every non-empty set satisfying the axioms of $\sf{ZF}$ be totally ordered? Let us first propose the following axiom,

Axiom of Ordering $(\sf{AO})$. If $S$ be a non-empty set. Let $a,b\in S$. Then the set $\{a,b\}$ can be totally ordered. 

Now let us consider $\sf{ZFO}$ set theory where $\sf{AC}$ is replaced by $\sf{AO}$. Then,

Can every non-empty set satisfying the axioms of $\sf{ZFO}$ be totally ordered?

By the principle of mathematical induction it follows that every finite set can be totally ordered. Intuitively it also seems to me that every infinite set can be totally ordered but I am unable to write a proof (assuming it exists). 
Can anyone help? 

Edit:- In view of the answer below, the question should be modified as,

Can every non-empty set satisfying the axioms of $\sf{ZF}$ be totally ordered?

 A: The answer is maybe negative. It depends on your universe of sets outside the model.

*

*There might not be any models of $\sf ZF$ to begin with. In that case there is a vacuous positive answer.


*The meta theory might include the axiom of choice, in which case every set can be ordered, in particular models of $\sf ZF$.


*But it could be that the answer is negative, too. If $\kappa$ is inaccessible, and we add a set which cannot be linearly ordered of rank $<\kappa$, then $V_\kappa$ still satisfies the axioms of $\sf ZF$, but it cannot be linearly ordered since a linear ordering will imply every set inside $V_\kappa$ can be ordered as well.
One can do even more, and by being careful make $V_\kappa$ non-linearly orderable, while at the same time not violate the axiom of choice below $\kappa$. So not even if you have a model of $\sf ZFC$ you cannot guarantee it to be linearly orderable.
A: $\operatorname{ZF}$ already proves $\operatorname{AO}$. Given two sets $a,b$ we have the total ordering $\preceq := \{ (a,a), (a,b), (b,b) \}$ on $\{a,b\}$.
What you probably have in mind is the following: Define a set of socks $S$ to be a set that consists entirely of pair $\{x,y\} \in S$. Let $\operatorname{AOS}$ state that there is a linear order on every set of socks.
We know that $\operatorname{ZF+ \neg AOS}$ is consistent relative to $\operatorname{ZF}$. I suspect that $\operatorname{ZF + AOS + \neg AC}$ is also consistent, but I'm not aware of a suitable model ( and I further conjecture that Asaf is).

Regarding the consistency of $\operatorname{ZF} \text{ + not every set has a linear order}$: It is indeed possible that there is a model of $\operatorname{ZF}$ which contains a set without a linear order. See the post over at mathoverflow for a reference and further discussions.
