Is $(X \setminus \operatorname{Cl}(A)) \cup A$ dense in $X$? The question I haven't been able to answer is the following:$\newcommand{\Cl}{\operatorname{Cl}}$
Is $(X \setminus \Cl(A)) \cup A$ dense in $X$?
To solve this problem I've tried the following:
First we note that a set $A$ is dense in another $X$ if and only of $\Cl(A)=X$ with this, we reduce our problem to proving that $\Cl((X\setminus\Cl(A))\cup A)=X$. For this, I've tried double inclusion. The one from left to right is fairly easy as $(X\setminus\Cl(A))\cup A$ is contained in X. The problem comes with the other inclusion, $x\in X \Rightarrow x\in \Cl((X\setminus\Cl(A))\cup A)$
First, to make things later on easier I've written down what means to for $x$ to be in $\Cl((X\setminus\Cl(A))\cup A)$:
$x\in \Cl((X\setminus\Cl(A))\cup A) \iff \forall U \in \tau \textrm{ with } x \in U \textrm{ then } U\cap ((X\setminus\Cl(A))\cup A) \neq \emptyset$
Now, playing around with this last part, I've done the following:
$U\cap ((X\setminus\Cl(A))\cup A) = (U\cap (X\setminus\Cl(A)))\cup(U\cap A) = (U\setminus\Cl(A))\cup(U\cap A) = \\ =((U\setminus\Cl(A))\cup U)\cap ((U\setminus\Cl(A))\cup A)=(U)\cap ((U\setminus\Cl(A))\cup A)$
With this I was trying to get to a point where I could easily see that this set is non empty for all $x$ in $X$ but I'm not able to finish this proof.
I'd appreciate some tips on how to get unstuck and on how to approach this kind of problems in general
Thanks in advance.
 A: 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.
A: 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 $X$.
