proving it is Hausdorff Question:
Let $X$ be a topological space and let $A ⊂ X$. Define an equivalence relation $∼$ on
$X$ such that the equivalence classes are:
• $A$ itself, and,
• all singletons $\{x\}$ such that $x \notin A$.
Then define $X/A$ to be the quotient space $X/{∼}$.
Prove that if $X$ is regular and $A$ is closed then $X/A$ is Hausdorff.
My Answer: 
The quotient mapping $q(x)$ is 
$$q(x)=\begin{cases}
x,&\text{if }x∉A\\
A,&\text{if }x∈A\;.
\end{cases}$$
Pick two distinct points $x,y∈X/A$.  If neither point is $A$, then $x,y∈$$X$ $\backslash$ $A$.  With the given information in the problem, particularly that $A⊂X$ is closed, we may deduce that $X$ $\backslash$ $A$ is open in $X$.  Then there exist open neighborhoods $U_A$ of $x$ and $V_A$ of $y$ such that $U_A$ and $V_A$ are disjoint from the closed set A.  Given that $X$ is regular, we can deduce that $X$ is Hausdorff.  $x$ and $y$ are distinct and thus we can find open sets $U'$ and $V'$ disjoint from each other such that $x∈U'$ and $y∈V'$.  Now define $U=U_A∩U′$ and $V=V_A∩V′$.  Then $U$ and $V$ are open, disjoint neighborhoods such that $x∈U$ and $y∈V$, and $U$, $V$ are also disjoint from $A$.
If one of the points, say $y$, is $A$, then $x$, the other distinct point, is $∈$ $X$ $\backslash$ $A$.  Because $X$ is regular, every non-empty closed subset of $X$,including  $A ⊂ X$, and every point in $X$ contained in a closed subset, admit non-overlapping open neighborhoods.  Then the point $x$ $∈$ $X$ $\backslash$ $A$ has an open neighborhood and the closed set A has a non-overlapping open neighborhood.  If $y$ is $A$, then $x$ and $y$ are contained in disjoint open neighborhoods.  
Therefore any two distinct points $x,y∈X/A$, whether or not they are $A$, have open neighborhoods $U$ of $x$ and $V$ of $y$ such that $U$ and $V$ are disjoint.
$X/A$ is Hausdorff. $□$
Is my proof even correct?
Please help me make my proof more rigorous and accurate.  I need everything to be absolutely clear and rigorous. Thank you.
 A: 
Pick two distinct points $x,y∈X/A$.  If neither point is $A$, then
  $x,y∈$$X$ $\backslash$ $A$.  With the given information in the
  problem, particularly that $A⊂X$ is closed, we may deduce that $X$
  $\backslash$ $A$ is open in $X$.  Then there exist open neighborhoods
  $U_A$ of $x$ and $V_A$ of $y$ such that $U_A$ and $V_A$ are disjoint
  from the closed set A.  Given that $X$ is regular, we can deduce that
  $X$ is Hausdorff.  $x$ and $y$ are distinct and thus we can find open
  sets $U'$ and $V'$ disjoint from each other such that $x∈U'$ and
  $y∈V'$.  Now define $U=U_A∩U′$ and $V=V_A∩V′$.  Then $U$ and $V$ are
  open, disjoint neighborhoods such that $x∈U$ and $y∈V$, and $U$, $V$
  are also disjoint from $A$.

So you take disjoint open neighborhoods $U'\ni x$ and $V'\ni y$, and you want them to also be disjoint from $A$. (Why is that sufficient in order to have disjoint neighborhoods of $\{x\}$ and $\{y\}$ in $X/A$? Here you could explain a bit more.) What you do is correct, but one could shorten it a bit by simply intersecting $U'$ and $V'$ with the open set $X/A$.

If one of the points, say $y$, is $A$, then $x$, the other distinct
  point, is $∈$ $X$ $\backslash$ $A$.  Because $X$ is regular, every
  non-empty closed subset of $X$, including  $A ⊂ X$, and every point in
  $X$ contained in a closed subset, admit non-overlapping open
  neighborhoods. Then the point $x$ $∈$ $X$ $\backslash$ $A$ has an
  open neighborhood and the closed set $A$ has a non-overlapping open
  neighborhood.  If $y$ is $A$, then $x$ and $y$ are contained in
  disjoint open neighborhoods.

What do you mean by "every point in $X$ contained in a closed subset"? The disjoint neighborhoods exist for every closed set $A$ and every point outside of $A$. Again, why does this show that there are disjoint open neighborhoods around $\{x\}$ and $A$ in $X/A$?
One more thing: It is bad style to write something like "then $x$, the other point, is $\in X/A$", as one should not mix text and symbols. You could write "then $x$, the other point, is in $X/A$" or "then $x\in X/A$"

Therefore any two distinct points $x,y∈X/A$, whether or not they are
  $A$, have open neighborhoods $U$ of $x$ and $V$ of $y$ such that $U$
  and $V$ are disjoint. $X/A$ is Hausdorff. $□$

