Proof that the axiom of choice implies the dense ordering principle

Question

The dense ordering principle sates that, for every infinite set X, there is a relation $\leq$ in X such that (X, $\leq$) is a dense linear order. I'm not sure if this idea of proof is correct: I sarted by stating that we can construct a choice function:

For X a countably infinite set (and therefore such that there is a bijection from X into the natural numbers).

f: P($X \times X$) $\rightarrow$ $\bigcup P(X)$ such that if Z $\subseteq$ Y $\in$ P($X \times X$), then f(Z) $\leq$ f(Y), and therefore the choice function f induces a linear order in $P(X \times X)$ (as I hope we can show).

Now we let A $\subseteq$ P($X \times X$) be the set of equivalence classes of $X \times X$ such that $[a,b]_≈= (c,d)$ and a/b= c/d.

Then, because P($X \times X$) can be linearly ordered, then its subset A can also be linearly ordered. Therefore, because A can be shown to be bijective to a subset of the rationals, then A is densely ordered.

Answer 1

One way is to use upward Lowenheim-Skolem / compactness theorem. Since the property of being a dense linear order is first-order axiomatizable, there are dense linear orders of every cardinality. Thus any set can be put in one to one correspondence with a dense linear order and thus has a dense linear order.

Answer 2

If you're allowed the full axiom of choice, well-order $X$ with $\prec$, and consider $(X\times\Bbb Q,\leq_{\rm Lex})$ as the lexicographic product of $\prec$ and $\leq_\Bbb Q$. Now use the axiom of choice again, and prove that $X\times\Bbb Q$ is the same size as $X$.