Given topological space $\{ (X_{i},O_{i}) \}_{i\in I}$, we can form their disjoint union $X := \amalg_{i}X_{i}$. Now if each $(X_{i},O_{i})$ satisfy the same separation axiom ($T_{0}, T_{1}, T_{2}, T_{2 \frac{1}{2}}, T_{3}, T_{3 \frac{1}{2}}, T_{4}, T_{5}, T_{6}$), I think so does their disjoint union
I think for $T_{6}$, we can use the pasting lemma(or by the universal construction). Given disjoint closed set $A,B \subset X$, for any $i$, $A_{i}:=A\cap X_{i}$ and $B_{i}:=B\cap X$ are both closed, so there exists continuous $f_{i}:X_{i} \to \mathbb{R}$ on $X_{i}$ that exactly separate $A_{i}$ and $B_{i}$, define $f:X \to \mathbb{R}$ such that $f: (i,x (\in X_{i})) \mapsto f_{i}(x)$, then by pasting lemma(or the universal construction of disjoint union) $f$ is continuous.
I think the other ones are true as well, but will be too much to write down.
So are all these separation axioms preserved under disjoint union? Perhaps there is a quick way to see that they all hold? Or maybe there's a counterexample?
I know there are wiki posts such that all separation axioms are preserved under homeomorphisms, and I hope this post can serve as a confirmation or to show a counterexample for future readers. (If it seems true, I will maybe write down all cases when I get a chance)