How do you prove that a class $\mathcal{A} $ of subsets of X which contains X, is closed under taking complements and is closed under taking finite disjoint unions is not necessarily an algebra?
$\mathcal{A}$ is an algebra if: (i) $\emptyset \in \mathcal{A}$, (ii) if $A, B \in \mathcal{A}$ then $A \cap B \in \mathcal{A}$ and (iii) if $A \in \mathcal{A}$ then $A^c \in \mathcal{A}$.
Counter-example: Let $X=\{1,2,3,4\}$. Let $A=\{1,2\}$, $B=\{2,3\}$, $C=\{3,4\}$, $D=\{1,4\}$. Let $\mathcal{A}=\{\emptyset,X,A,B,C,D\}$. It can be proved that $\mathcal{A}$ satisfies all the stated conditions. However, $A\cap B=\{2\}\notin\mathcal{A}$.