A property in Hausdorff compact spaces

Let $$X$$ be a Hausdorff compact space and $$A$$ and $$B$$ be two disjoint closed subsets of $$X$$.

Since $$X$$ is Hausdorff and compact, it is normal, hence there exists $$A_1$$ and $$B_1$$ disjoint neighborhoods of $$A$$ and $$B$$. Since $$X$$ is Hausdorff, $$A_1$$ and $$B_1$$ can be chosen s.t. $$\overline{A_1}\cap\overline{B_1} =\emptyset.$$

In this way we can construct family of neighborhoods $$\{ A_n \}$$ and $$\{ B_n\}$$ for $$A$$ and $$B$$ with the properties $$\overline{A_n} \subseteq A_{n+1},\; \overline{B_n} \subseteq B_{n+1}$$ and $$\overline{A_n}\cap \overline{B_n}=\emptyset,\; \forall n\in\mathbb{N}.$$

My question: Is there a way to construct $$\{ A_n \}$$ and $$\{ B_n\}$$ s.t. $$X=\bigcup\limits_{n\in\mathbb{N}} A_n \cup \bigcup\limits_{n\in\mathbb{N}} B_n$$?

This is not true. Let $$X=[0,1], A=\{0\}, B=\{1\}$$, $$A_n-[0,\frac 1 3 -\frac 1n), B_n=(\frac 2 3-\frac 1 n,1]$$. Then $$\cup A _n \cup \cup B_n \neq X$$.
Answer for the revised verion of the question: This cannot be done if $$X$$ is connected. If $$X=\cup A _n \cup \cup B_n$$ then there is a finite subcover for this open cover. In view of monotonicity of the sets this gives $$X=A_N \cup B_N$$ for some $$N$$. But this gives a separation of $$X$$.