Proving that ${\left( {A \cup B} \right)^{\text{o}}} = \emptyset$ given that ${\left( {\bar A} \right)^{\text{o}}} = \emptyset = {B^{\text{o}}}$ I have the following question:

Let $X$ be a topological space and let $A,B \subseteq X$ such that ${\left( {\bar A} \right)^{\text{o}}} = \emptyset  = {B^{\text{o}}}$. Show that ${\left( {A \cup B} \right)^{\text{o}}} = \emptyset$.

 Am I allowed to say that the empty set is a subset of every set so $\emptyset  \subseteq {\left( {A \cup B} \right)^{\text{o}}}$?
 I'm struggling with the reverse inclusion: I have the following $ {\left( {A \cup B} \right)^{\text{o}}} \subseteq A \cup B \subseteq \overline {A \cup B}  = \bar A \cup \bar B\ $, but I'm not sure where to go from there?
 A: Let $U\subseteq A\cup B$ be open.
Then $V=U\setminus \bar A$ is open and $V\subseteq (A\cup B)\setminus \bar A\subseteq B$, so $V=\emptyset$. Thus, $U\subseteq \bar A$, whence $U=\emptyset$.
Note that both conditions are necessary: if $A$ is any set such that $\bar A$ has nonempty interior, then $B=\bar A\setminus A$ will have empty interior and $A\cup B=\bar A$ will not.
A: $ {\left( {\overline A} \right)^{\text{o}}}=\emptyset  \iff {\left ( {\left( {\overline A} \right)^{\text{o}}}  \right)}^c=X$ $ \iff   {\overline {{(A^c)^{\text{o}}}}}=X$.
Then ${(A^c)^{\text{o}}}$ is dense in $X$ or equivalently ${(A^c)^{\text{o}}} \cap \theta \neq \emptyset$ for all non-empty open sets $\theta$ from $X$.
$(A \cup B)^{\text{o}} \cap {(A^c)^{\text{o}}}={\left[ (A \cup B) \cap (A^c) \right]}^{\text{o}}={\left[ B \cap (A^c) \right]}^{\text{o}} \subset B^{\text{o}}=\emptyset$.  
Since $(A \cup B)^{\text{o}}$ is open set, it must be empty.
Hence  $(A \cup B)^{\text{o}}= \emptyset$.
A: Suppose that $U$ is open, non-empty and $U \subseteq A \cup B$.
Clearly, $U \subseteq \overline{A} \cup B$ as well, and as $\overline{A}$ is closed, $U \setminus \overline{A} = U \cap \overline{A}^\complement$ is open too and must be non-empty (as otherwise $U \subseteq \overline{A}$ which doesn't hold as the interior of $\overline{A}$ is empty!), and by the inclusion we have $U\setminus \overline{A} \subseteq (\overline{A} \cup B)\setminus \overline{A} \subseteq B$, which contradicts $B^\circ=\emptyset$. So no such non-empty $U$ can exist and so $(A \cup B)^\circ = \emptyset$.
A: We prove for general case $A=\bar{A}$ since $(A \cap B)^o \subseteq (\bar{A} \cap B )^o$. I use the idea that $A^o= \emptyset$ iff $X=(A^o)^c=\overline{A^c}$
So, we need to prove that $\overline{(A\cup B)^c}=\overline{A^c \cap B^c}=X$ for $A,B \subseteq X$ such that $A$ closed and $A^o=B^o=\emptyset$.
Take any $x \in X$ and any open neighborhood $U$ of $x$. We need to prove that $U \cap A^c \cap B^c \neq \emptyset$. 
Since $A^o= \emptyset$ so $\overline{A^c}=X$. Therefore, $U \cap A^c \neq \emptyset$. Take $y \in U \cap A^c$. Since $A$ is closed, $A^c$ is open, there exist an open neighborhood $V$ of $y$ such that $y \in V \subseteq U \cap A^c$. 
Since $B^o = \emptyset$ so $\overline{B^c}=X$. Therefore, for open neighborhood $V$ of $y \in X$, $V\cap B^c \neq \emptyset$. So $U \cap A^c \cap B^c \supset V\cap B^c \neq \emptyset$. 
Another way if you're not familiar with the above idea.
By contradiction, assume that there exists $x \in (A \cup B)^o$.
So there must be an open neighborhood $U$ of $x$ such that $ x \in U \subset A \cup B$.


*

*$1^{st}$ case, $ x \in U \subset A$, contradiction to $A^o =\emptyset$.

*$2 ^{nd}$ case, $ x \in U \not\subset A$ there exists $ y \in U$ and $y \notin A$ so $y \in B$. Since $A$ closed, $A^c$ open, so there exists open neighborhood $V$ of $y$ such that $ y \in V \subset A^c$. So $y \in V \cap U \subset U \subset A \cap B$. Note that $ V \cap U$ is open and $V\cap U \subset V \in A^c$, so $y \in V\cap U \in B$, which contracdicts to $B^o =\emptyset$.  

