Is $(X \setminus \operatorname{Cl}(A)) \cup A$ dense in $X$?

Just note that

$\overline{(X\setminus \overline{A}) \cup A} = \overline{X\setminus \overline{A}} \cup \overline{A} \supseteq (X\setminus \overline{A}) \cup \overline{A}= X$ so your set is dense. We use that finite unions commute with closure and the closure only gets bigger.

We use extensively the fact that a point $x$ is in the closure of $K$ if and only if all open sets containing $x$ have a non-empty intersection with $K$.

Choose $x \in X$. If $x \notin \overline A$, then $x \in X \setminus \overline A$, so any open set $U$ containing $x$ necessarily intersects $X \setminus \overline A$ (if nothing else, the intersection includes $x$) and $U \cap ((X \setminus \overline A) \cup A) \supseteq U \cap (X \setminus \overline A) \neq \varnothing$.

If $x \in \overline A$, then because $x$ is in the closure of $A$, we know that any open set $U$ containing $x$ also has non-empty intersection with $A$; i.e., $U \cap A \neq \varnothing$. Thus, again in this case, $U \cap ((X \setminus \overline A) \cup A) \supseteq U \cap A \neq \varnothing$.

Together, these two statements prove that any open set containing any point of $X$ must have non-empty intersection with $(X \setminus \overline A) \cup A$, so $X$ is the closure of $(X \setminus \overline A) \cup A$, which is therefore dense in $X$.