Let's assume that $\mathcal{A}$ is an algebra over $\mathit{X}$, and $\mathcal{A}$ is closed under countable intersections of descending set sequences; if $A_1 \supseteq A_2 \supseteq A_3 \supseteq ... $ and $A_i \in \mathcal{A}, i \in \mathbb{N}$, then $\bigcap\limits^{\infty}_{i=1}A_i \in \mathcal{A}$. Prove that $\mathcal{A}$ is $\sigma$-algebra on $X$.
My attempt: Let $\forall n \in \mathbb{N} \; B_n \in \mathcal{A}$. Then define $A_n = B_n \setminus \bigcup\limits_{i=1}^{n-1} B_i$. Then $A_n$ is descending family of sets, $A_n \in \mathcal{A}$ as this is algebra. We know that $\bigcap\limits^{\infty}_{i=1}A_i \in \mathcal{A} \leftrightarrow \bigcap\limits^{\infty}_{i=1}(B_n \setminus \bigcup\limits_{i=1}^{n-1} B_i) \in \mathcal{A}$. Now, I was thinking about using De Morgan's laws, etc., but I can't obtain the result and I don't know if I haven't messed up in the beginning.