Let $A\subseteq\mathbb{R}$ with $A$ uncountable.
We define $\prec$ on $A\times A$ by $(a,b)\preceq (c,d)\iff a\leq c\land b\leq d$.
Then $x\prec y\iff x\preceq y\land x\neq y$.
Now prove there's no partition of $A\times A$ of countably many $X_i$'s where $(X_i,\prec )$ is total for every $i$.
I was thinking of somehow deriving a contradiction by finding an uncountable number of rationals or an embedding into $\omega_1$.
This is also similar to this question: Prove there's no partition of $\omega_1\times \omega_1$ with the following property
But the same argument doesn't work as there can be two uncountable subsets of $\mathbb{R}$ with one above the other.
Let $X \subset A\times A$ be a subset so that $\prec$ restricted to $X$ is total. For $x\in A$, put $X^x=\{y\in A: (x,y)\in X\}$. Note that if $x \neq x'$ and $y_1,y_2$ are two distinct points in $X^x$ and similarly $y_1', y_2'$ are two distinct points in $X^{x'}$, then since $(X, \prec)$ is totally ordered, the two intervals $[y_1, y_2]$ and $[y_1', y_2']$ are disjoint. Hence there are only countably many $x\in A$ for which $X^x$ is not a singleton. Indeed if there were uncountably many such $x$'s we get a an uncountable pairwise disjoint family of non-degenerate intervals which is impossible.
Now, if $\{X_n\}_n$ is an countable collection of such totally ordered sets we apply the above to see that there must be an $x\in A$ such that $X_n^x$ is a singleton for all $n$. But this surely means that the $X_n$s cannot cover $\{x\}\times A$.