Let $A_n$ and $B_n$ be two sequences of sets. How $(\liminf_n A_n \cup \liminf_n B_n)$ and $\liminf_n (A_n\cup B_n)$ are related?
Def. Given a sequence of sets $E_n$, the limit inferior of $E_n$ is defined as $$\liminf_{n\to\infty} E_n=\bigcup_{n=1}^\infty \bigcap_{k=n}^\infty E_k$$
Some thoughts
Write $\liminf_n A_n=\bigcup_{n}C_n$ and $\liminf_n B_n=\bigcup_{n}D_n$ where $C_n=\bigcap_{k=n}^\infty A_k$ and $D_n=\bigcap_{k=n}^\infty B_k$.
I will use a (intutive) result that requires a proof: $(\bigcup_{n\in\mathbb{N}}C_n) \cup (\bigcup_{l\in\mathbb{N}}D_l)=\bigcup_{n\in\mathbb{N}}C_n\cup D_n$.
On the other hand, for each $n$, $$C_n\cup D_n=\bigcap_{k=n}^\infty A_k \cup \bigcap_{l=n}^\infty B_l=\bigcap_{k=n}^\infty \left[ A_k \cup \left(\bigcap_{l=n}^\infty B_l \right)\right]\subseteq \bigcap_{k=n}^\infty A_k \cup B_k.$$
From these observations, we immediately have
$$\liminf_n (A_n\cup B_n)\supseteq \liminf_n A_n \cup \liminf_n B_n $$
Best Answer
We go to show that: $\liminf_{n}A_{n}\cup\liminf_{n}B_{n}\subseteq\liminf_{n}(A_{n}\cup B_{n})$.
Let $x\in LHS$, then $x\in\liminf_{n}A_{n}$ or $x\in\liminf_{n}B_{n}$. Suppose that $x\in\liminf_{n}A_{n}$, then there exists $n$ such that $x\in\cap_{k\geq n}A_{k}$. For each $k\geq n$, $x\in A_{k}\Rightarrow x\in A_{k}\cup B_{k}$. Therefore, $x\in\cap_{k\geq n}(A_{k}\cup B_{k})$. Hence $x\in\cup_{n}\cap_{k\geq n}(A_{k}\cup B_{k})=RHS$. Similarly, if $x\in\liminf_{n}B_{n}$, we can show that $x\in RHS$. This shows that $LHS\subseteq RHS$.