Let $X$ be a $\sigma$-compact topological space, meaning that $X = \bigcup_{i\in\mathbb{N}} K_i$ for a sequence of compact subsets $(K_i)$.
Does this already imply that $X$ can be written as a union
$$\tag{1} X \,=\, A\cup B \qquad\text{with} \qquad A \,=\, \bigcup_j K^{a}_j \quad\text{and}\quad B \,=\, \bigcup_k K^{b}_k$$
where each $(K_\nu^{\gamma})_\nu$ [$\gamma\in\{a, b\}$] is a sequence of pairwise disjoint compact sets?
Let $X_i$ be a sequence of distinct uncountable sets, $x$ a point not in any $X_i$, and $X = \bigcup\limits_{i = 1}^\infty X_i \cup \{x\}$. Say that $Y \subseteq X$ is open if it either doesn't contain $x$, or it contains $x$ and $X_i \setminus Y$ is finite for all $i$ - essentially take countably many copies of uncountable space with discrete topology, take Alexandroff extension of each, and glue them by the extension point.
Each $X_i \cup \{x\}$ is compact: any open cover of it includes some neighbourhood of $x$, which also includes all but finitely many points from $X_i$, thus we can choose finite subcover. So, $X$ is $\sigma$-compact.
But all compact subsets of $X$ include points from only finitely many $X_i$: if $Y$ includes points from infinitely many $X_i$, then let $Z \subset Y$ such that $Z$ is infinite and intersects infinitely many $X_i$ in $1$ point, then $\{z_1\}, \{z_2\}, \ldots, X \setminus Z$ is open cover of $Y$ that has no finite subcover. Moreover, any compact subset that doesn't include $x$ is finite.
So, if we represent $X$ as union of compact sets, some of them are finite, and the rest include $x$ together with points from finitely many $X_i$. If our union is countable, then, as every $X_i$ is uncountable, every $X_i$ has points from some compact set containing $x$. But as every compact set containing $x$ intersects only finitely many $X_i$, and there are infinitely many of them, there is infinite amount of compact sets containing $x$ in the union.
Thus any representation of $X$ as countable union of compact sets includes infinitely many distinct sets that include $x$, and thus can't be split into finitely many sequences of sets where all sets from one sequence are pairwise disjoint.