I need to prove that if $X/A$ is Hausdorff, then $A$ is closed. I have proven the converse (adding the condition that $X$ is regular), which I have found floating around, but I was unable to find any hints for this one.
My attempt: (Note that I am using $\pi$ as the quotient map) $\pi(A)$ is a single point, so it is compact, and thus closed. Since $\pi$ is continuous, $\pi^{-1}(\pi(A))=A$ is also closed.
That last line is the one I'm not convinced of, not to mention that this seems too easy an argument. Any tips would be appreciated, trying to go through all my confusing questions before finals!
The idea is correct. Letting $\pi$ be the quotient map, we know $X{/}A$ is Hausdorff so $T_1$, and particular all singletons are closed. It follows that $\{\pi(a)\}$ for some $a \in A$ (or "$A$ as a point" of the quotient) is closed and indeed by the definition of the identification $A=\pi^{-1}[\{\pi(a)\}]$ and thus $A$ is closed as the inverse image of a closed set under a continuous map. Compactness of $\{\pi(a)\}$ is overkill (and needs another stronger theorem (a compact subset of a Hausdorff space is closed) than plain Hausdorff impying $T_1$ which is easy).
Indeed if $X$ is regular and $A$ is closed we have that $X{/}A$ is Hausdorff by an obvious argument; separating $A$ as a point and some other $x$ in this quotient depends on being able to separate $A$ (as a set) from $x$ in $X$.