Union of two $\sigma$-algebras is not $\sigma$-algebra

take $X := \{a,b,c\}$ and $A_1 := \{ \{a\}, \{b,c\}, \emptyset, X\}$, $A_2 := \{ \{b\}, \{a,c\}, \emptyset, X\}$ and show that $A_1 \cup A_2$ is not a $\sigma$-algebra

Let it be that $X=A\cup B=U\cup V$ with $A\cap B=\emptyset=U\cap V$. Then $\mathcal{A}=\left\{ \emptyset,A,B,X\right\} $ and $\mathcal{V}=\left\{ \emptyset,U,V,X\right\} $ are both $\sigma$-algebras. Is $\mathcal{A}\cup\mathcal{V}$ a $\sigma$-algebra?

Not if $A\cup U\notin\mathcal{A}$ and $A\cup U\notin\mathcal{V}$.