Proving a function $u:[0,\infty)^2 \to \mathbb{R}$ cannot preserve an 'order' given to the domain The set $[0,\infty)^2$ is considered to have order $\succeq$, where $\succeq$ is a binary relation satisfying $(x_1,y_1) \succeq (x_2,y_2)$ if $x_1 > x_2$ or ( $x_1 = x_2$ and $y_1 \geq y_2$ ). I am trying to show no function $u:[0,\infty)^2 \to \mathbb{R}$ can preserve this order (strictly), with the codomain ordered in the obvious way.
I would like to suppose it exists for a contradiction and argue that, for $\omega_{x} = (x,1)$ and $\omega_{x}' = (x,2)$, it is necessarily the case that $u(\omega_x') > u(\omega_x)$ for any $x \in [1,2]$, and hence as for each such value $x$ the value of $u$ must increase, and with $x$ belonging to an uncountable set and each increase is strictly positive, and $u$ is strictly monotonic in the orderings, indeed $u(\omega_2)$ would have to be infinitely bigger than $u(\omega_1)$.
However, I am struggling to formulate this answer properly. The argument relies on the cardinality of the reals being uncountable, however it is exactly this that is causing me difficulty in formulating a proof.
How can I go about formulating this idea into a rigorous proof? If possible, I'd like this to not rely on advanced mathematical structures much further than what I have already mentioned.
 A: I am assuming that the relation $\{(p,q)\in [0,\infty)^2: p\geq q\ne p\}$ is a linear order.  Suppose $f:[0,\infty)^2\to \Bbb R$  strictly preserves this linear order.  
For each $x\in [0,\infty)$ let $S(x)$ be the non-empty real interval $(f(x,0),f(x,1)).$ 
If $x,x'\in [0,\infty)$ with $x< x'$ then $S(x)\cap S(x')=\phi$ because  $f(x,1)<f(x',0).$ 
So $\{S(x): x\in [0,\infty)\}$ is an uncountable family of pair-wise disjoint non-empty open subsets of $\Bbb R.$ So $\{S(x)\cap \Bbb Q: x\in [0,\infty)\}$ is an uncountable family of pair-wise disjoint non-empty subsets of $\Bbb Q$, which is impossible because  $\Bbb Q$ is only countable.
A: Pay attention to each vertical line $\{t\}\times[0,\infty)\subseteq[0,\infty)^2$. Each of these vertical lines inherits the order from $[0,\infty)^2$, and the order structure on each vertical line is isomorphic to $[0,\infty)$.
Now suppose for contradiction there is a function $u$ preserving the order strictly. Note that we require $u(\{t\}\times[0,\infty))$ is bounded in $\Bbb R$, i.e. $u(\{t\}\times[0,\infty))\subseteq[a_t,b_t]$ for some $a_t,b_t\in\Bbb R$ depending on $t$. To see why, notice that $u(\{t\}\times[0,\infty))$ is bounded from below because there is a smallest element $(t,0)$ on the vertical line, and is bounded from above because each element in $u(\{t\}\times[0,\infty))$ has to be less than $u(t+1,0)$. Note that you can take $a_t=u(t,0)$. Note also $b_t\gt a_t$ because the increasing-ness is strict. With no risk, pick $b_t=\sup u(\{t\}\times[0,\infty))$.
Let $\varepsilon\gt0$. Since $a_{t+\varepsilon}$ is an upper bound for the set $u(\{t\}\times[0,\infty))$, by definition of supremum, we have $a_{t+\varepsilon}\ge b_t$. So we obtain an uncountable number of closed, bounded and nondegenerate intervals $\{[a_t,b_t]:t\in[0,\infty)\}$ which are distinct from each other in the sense $[a_t,b_t]\not=[a_{t'},b_{t'}]$ whenever $t\not=t'$ due to the relation $a_{t+\varepsilon}\ge b_t$.
Now since those intervals are nondegenerate, we can pick a rational number $r_t$ with $a_t\lt r_t\lt b_t$ for each $t$. This requires axiom of choice. Again, by the relation $a_{t+\varepsilon}\ge b_t$, the $r_t$'s are distinct in the sense $r_t\not=r_{t'}$ whenever $t\not=t'$. We have found an uncountable number of rational numbers $\{r_t:t\in[0,\infty)\}$, a contradiction.
