I have the following argument which I encountered, and can't seem to find why it's not true:
Let $X$ be a topological space and let $A$ and $B$ be nonempty subsets of $X$. Then $\overline{A\cap B}^A=\overline{A\cap B}^X\cap A$.
"Proof"
$\overline{A\cap B}^A =\cap \{ F\subset A: \; F\supseteq A\cap B, F \text{ is closed in } A \}$.
Since $F$ is closed in $A$ if and only if $F=F'\cap A$ where $F$ is closed in $X$. Hence
$\overline{A\cap B}^A= \cap \{ F'\cap A: \; F'\supseteq A\cap B, F' \text{ is closed }\}= \Big( \cap \{ F': \; F'\supseteq A\cap B, F' \text{ is closed }\}\Big) \cap A $.
And using an equivalent characterization of the closure,
$ \cap \{ F': \; F'\supseteq A\cap B, F' \text{ is closed }\}=\overline{A\cap B} $.
And finally we conclude
$ \overline{A\cap B}^A=\overline{A\cap B}^X\cap A $.
I can't seem to find where this argument falters, but this thread makes me think that it's wrong, and would appreciate any pointers.
I am not sure whether this answers your question but in general if $D\subseteq A$ then $\overline D^A=A\cap\overline D$.
Proof:
Observe that $A\cap\overline D$ is closed in $A$ with $D\subseteq A\cap\overline D$ telling us that $\overline D^A\subseteq A\cap\overline D$.
Conversely $\overline D^A$ is closed in $A$ so that $\overline D^A=A\cap F$ for some closed set $F$.
Then from $D\subseteq\overline D^A$ it follows that $D\subseteq F$ and consequently $\overline D\subseteq F$.
Then $A\cap\overline D\subseteq A\cap F=\overline D^A$.
Applying this on $D:=A\cap B$ we find the statement in your question.