Suppose $X$ is a discrete space and $A$ and $B$ are disjoint subspaces of $X$.
Then consider $\beta(X) \times \beta(X)$, We can think of $A \times A, B \times B$ as a subspace of $ \beta(X) \times \beta(X)$, where $\beta(X)$ is Stone-Cech compactification of $X$.
Then it’s not true that $\overline{A \times A} \cap \overline{B \times B}$ is empty. So $\overline{A} \cap \overline{B}$ is non-empty.
But in Munkres’ Introduction to topology,
There’s an exercise which says that
If $X$ is discrete space and $A$ is a subspace of $X$, then show that $\overline{A}$ and $\overline{X \setminus A}$ are disjoint.
It seems to me that those are contradicting to each other. Is there any missing point I couldn’t see? Any comment will be appreciated.
Closures are taken in the (product of) compactification, not in the original $X$. And the compactification is not discrete. Hence no contradiction with Munkres.
The fundamental reason is that if $A\subseteq B\subseteq C$ then the closure of $A$ in $B$ need not be equal the closure of $A$ in $C$. In fact the latter may be bigger. To visualize this consider $X=\{0, 1\}$ with the antidiscrete topology $\{\emptyset, X\}$ and $A=B=\{0\}$. Note that $A,B$ are discrete and so the closure of $A$ in $B$ is just $A$. But the closure of $A$ in $X$ is whole $X$.
The situation when the intermediate closure is equal to the outer closure is when the middle subspace $B$ is closed in $X$. In your situation that would require $X$ to be closed in $\beta(X)$ and this can only happen when $X$ is compact to begin with (and so it is equal to its own Stone-Cech compactification).