The Question:
Let $X$ be a regular topological space and $A \subset X$. Let $Y = X/A$ be the quotient space defined by the equivalence relation on $X$ given by $x_1$ is equivalent to $x_2$ iff $x_1 = x_2$ or $x_1,x_2 \in A$. Prove that $Y$ is Hausdorff iff $A$ is closed.
I have read similar questions here and here. However, I feel these questions are inherently different, as they are asking about the relation being closed in $X \times X$. I am sure these statements are related, but I want to try and prove it without referring to the relation being closed.
My attempt
$(\Rightarrow)$ Suppose that $Y$ is Hausdorff. Consider $x \in X \backslash A$ and $y \in A$. Then $q(x) = [x] (\ne [A])$ and $q(y) = [y] = [A]$. Because $Y$ is Hausdorff, we have that there exist disjoint open sets $U,V \subset Y$ such that $[x] \in U$ and $[y] = [A] \in V$. Because $q$ is continuous, we have that $q^{-1}(U),q^{-1}(V) \in \mathcal{T}_X$. Furthermore, $x \in q^{-1}(U)$, $A \subset q^{-1}(V)$, and $q^{-1}(U) \cap q^{-1}(V) = \emptyset$. Hence, we have found an open neighborhood of $x$ that does not intersect $A$. So $X \backslash A$ is open, and $A$ is closed.
I think the above is fine. While it might be the cleanest proof, I think it is correct. My trouble is with the reverse implicaiton:
$(\Leftarrow)$ Suppose that $A$ is closed. Let $[x],[y] \in Y$ be such that $[x] \ne [y]$. First, consider the case when $[x] \ne [A] \ne [y]$. Then $q^{-1}([x]) = \{x\}$ and $q^{-1}([y]) = \{y\}$. Because $X$ is regular, we have that there exist disjoint sets $U,V$ such that $x \in U$ and $y \in V$.
Here I want to say that $q(U)$ and $q(V)$ are the desired sets in $Y$, but I know that $q$ is not necessarily an open map. Note: we can come up with similar sets in the case $[x] = [A]$ because $X$ is regular.
Where do I go from here? How do we construct the desired open sets?
