Is there a clever general way of disconnecting those sets? Let $X$ be a normal space. For the sake of brevity, I say that set $A$ "connects" $B$ with $C$ iff there is a connected set $\ell \subseteq A$ such that $\ell \cap B \neq \emptyset \neq \ell \cap C$.
Let $A_1,A_2,B_1,B_2$ be closed sets. Let $C_1,C_2$ be open sets. Suppose that:

*

*$A_1 \cap B_1 = \emptyset$

*$A_2 \cap B_2 = \emptyset$

*$C_1 \cup C_2 = X$

*$C_2$ doesn't connect $A_1$ with $B_1$

*$C_1$ doesn't connect $A_2$ with $B_2$
I wonder if there is some general way of using $C_1,C_2$ to construct open sets $D_1,D_2$ such that:

*

*$D_1 \cup D_2 = X$

*Neither $D_1$ nor $D_2$ connects $A_1$ with $B_1$

*Neither $D_1$ nor $D_2$ connects $A_2$ with $B_2$
It's quite intuitive to me, and it looks like it can be done with some clever application of Urysohn's lemma. But loooks and intuition are often deceiving, and so far I don't feel like I have mady any real progress and my mind is completely blank.
 A: Here is a proof from stronger versions of your hypotheses (4) and (5).  Specifically, let's assume that actually $A_1\cup B_1\subseteq C_2$ and that $A_1$ and $B_1$ can be separated by clopen sets in $C_2$, and similarly for $C_1$.
Let $E_1=X\setminus C_2$ and $E_2=X\setminus C_1$, so $E_1$ and $E_2$ are disjoint closed sets that each $E_i$ is disjoint from $A_i$ and $B_i$, and $A_i$ and $B_i$ are separated by clopen sets in $X\setminus E_i$.  We may enlarge the $E_i$ to assume they are in fact zero sets (take a continuous function that separates $E_1$ from $E_2\cup A_1\cup B_1$ to enlarge $E_1$ to a $G_\delta$ set with the same properties, and then do the same to enlarge $E_2$).  Let $U_i$ and $V_i$ be complementary clopen subsets of $X\setminus E_i$ which contain $A_i$ and $B_i$, respectively.  Now we can find a continuous function $f_i:X\setminus V_i\to[0,1]$ such that $f_i$ is $1$ on $A_i$ and $f_i^{-1}(\{0\})=E_i$ (the latter being possible since $E_i$ is a zero set).  Similarly, we can find $g_i:X\setminus U_i\to[-1,0]$ such that $g_i$ is $-1$ on $B_i$ and $g_i^{-1}(\{0\})=E_i$.  By the pasting lemma, $f_i$ and $g_i$ combine to give a continuous $h_i:X\to[-1,1]$ such that $h_i$ is $1$ on $A_i$, $-1$ on $B_i$, and $h_i^{-1}(\{0\})=E_i$.
Now define $h:X\to[-1,1]^2$ by $h(x)=(h_1(x),h_2(x))$.  This $h$ maps $A_1$ and $B_1$ to the vertical sides of the square and $A_2$ and $B_2$ to the horizontal sides of the square.  Moreover, $h^{-1}(\{(0,0)\})=E_1\cap E_2=\emptyset$.  Composing with a retraction from $[-1,1]^2\setminus\{(0,0)\}$ to the boundary of the square, we get a map $X\to \partial[-1,1]^2$ which maps $A_1$ and $B_1$ to the vertical sides of the square and $A_2$ and $B_2$ to the horizontal sides.  It thus suffices to solve the problem in the special case where $X=\partial[-1,1]^2$ and the $A_i$ and $B_i$ are the sides of the square, in which case it is easy.

More generally, the final paragraph above shows that it suffices to find a non-surjective map $h:X\to[-1,1]^2$ that maps the $A_i$ and $B_i$ to the sides of the square.  However, your hypotheses (even with the additional requirement that $A_1\cup B_1\subseteq C_2$ and $A_2\cup B_2\subseteq C_1$) do not suffice to prove the existence of such an $h$, so this method cannot solve your problem in full generality.  For instance, let $$X=[-1,1]^2\times\{1,1/2,1/3,\dots\}\cup\partial[-1,1]^2\times\{0\}\subset\mathbb{R}^3,$$ and let the $A_i$ and $B_i$ be the sides of the square $\partial[-1,1]^2\times\{0\}$.  If you let $C_1$ consist of the complement of a non-vertex point from each of $A_1$ and $B_1$ and similarly for $C_2$, then this satisfies all your hypotheses, and additionally $A_1\cup B_1\subseteq C_2$ and $A_2\cup B_2\subseteq C_1$.  However, no non-surjective map $h:X\to [-1,1]^2$ mapping the $A_i$ and $B_i$ to the sides of the square exists.  Indeed, if $h$ maps the $A_i$ and $B_i$ to the sides of the square, then it must have degree $1$ when restricted to a map $\partial[-1,1]^2\times\{0\}\to[-1,1]^2\setminus\{(0,0)\}$.  It then follows that $h$ is also has degree $1$ when restricted to $\partial[-1,1]^2\times\{1/n\}\to[-1,1]^2\setminus\{(0,0)\}$ for all sufficiently large $n$.  But then any extension of this degree $1$ map to all of $[-1,1]^2\times\{1/n\}$ will have to surject onto $[-1,1]^2$, so $h$ must be surjective.
(Note though that this example is not a counterexample to your question, since the desired $D_i$ exist--you can just take them to be complements of appropriate finite subsets of $\partial[-1,1]^2\times\{0\}$.  At this point my intuition is that the answer to your question is no in general, but I haven't been able to come up with a counterexample.  Mainly, the "not connecting" hypotheses seem rather hard to use in cases like the example above where components and quasicomponents are not the same.)
