Let $X$ be a Hausdorff compact space and let $A$ and $B$ be two disjoint closed sets in $X$.
Suppose that for any $z\in X$ we can find a neighborhood $U_z$ such that $U_z\cap A=\emptyset$ or $U_z\cap B=\emptyset$. Since $\{U_z\mid z\in X\}$ is a cover of $X$, we can find $\{z_1,z_2,\dots,z_n\}\subseteq X$ such that $\bigcup\limits_{k=1}^nU_{z_k}=X$.
A chain from $x$ to $y$ on $X$ is a family of open sets $\{U_k\}_{k=1}^n$ such that $x\in U_1,$ $y\in U_n$ and $U_k\cap U_{k+1}\neq\emptyset$, $\forall k\in\{1,2,\dots,n-1\}$.
I am asking if we can construct such a cover $\{U_{z_k}\}_{k=1}^n$ so that for $a\in A$ and $b\in B$ there is no chain made by elements of $\{U_{z_k}\}_{k=1}^n$ from $a$ to $b$?
P.S. I haven't used the Hausdorff condition yet!
Update:
Let me explain my whole purpose of the question: It is a well known fact that if $Q$ is a quasi-component in a Hausdorff compact space, then it must be connected. I want to prove this using the concept of chains.
There is a result that states: For $x\in X$, the set $Q(x)=\{y\in X|$ for every possible cover of $X$ there is a chain from $x$ to $y$ consisting of elements of the cover $\}$ is a quasi-component of $x$.
I want to prove that if $Q=A\cup B$ where $A$ and $B$ are disjoined closed sets, then there must exists at least one $z\in X$ s.t. every neighborhood of $z$ has nonempty intersection with both $A$ and $B$, and since $A$ and $B$ were closed they cannot be disjoined, which will contradict the assumption that $Q$ is disconnected.
So, I suppose that if $\forall z\in X$ there is a neighborhood s.t. does not intersect both $A$ and $B$, then there must be a cover of $X$ for which you cannot find a chain from an element in $A$ to an element in $B$.
Let $X$ be connected and $A, B$ nonempty. Assume such a cover $(U_{z_k})_{k=1}^n$ exists where no chain between $A$ and $B$. Then define $S$ to be the set of all indices $k$ such that there exists a chain to $A$. Furthermore, we define $$M=\bigcup_{k\in S} U_{z_k}$$ and $$ N= \bigcup_{k\in S^c} U_{z_k}.$$ By construction we get that $M, N$ are disjoint and open. As $A,B$ are nonempty, so are $M, N$. However, then $M$ is clopen, nonempty and different from $X$ which contradicts that $X$ is connected.