Show that if $X$ is $T_3 \implies$ $X/ \sim$ is Hausdorff Let $X$ be an Hausdorff space and $C \subset X$ closed.Let's define an equivalence relation $\sim$ such that two elements $x,y \in X$ are in relation $x \sim y \iff x=y$ or $x,y \in C$.This equivalence relation induces a quotient space $X/C$.
$1)$Show that $X/\sim$ is $T_1$;
$2)$Show that if $X$ is $T_3 \implies$ $X/ \sim$ is Hausdorff;
For the first part I noticed that $U$ is open in the quotien topology iff $U$ is an open saturated in $X$ topolgy.
So, let's consider $\pi: X \to X/\sim$ and we observe that $\forall x \in C \subset X$, $\pi^{-1}(\pi (x))=\pi^{-1}([x])=C$, while if $y \in X-C$, $\pi^{-1}(\pi (y))=\pi^{-1}([y])=y \implies \forall A\subset X-C$, $ \pi^{-1}(\pi (A))=\pi^{-1}([A])=A$.
As consequence, $[x] \neq [y]\iff x\neq y$ in $X-C$ or if $x \in C, y \in X-C$.
$X-C \ni x\neq y \in X-C \implies (\square)$ $\exists U,V\in \mathcal{T}_X:x\in U,y \in V$, with $U \cap V= \emptyset$ and $U,V \subset X-C$. This implies that $[y]\notin\pi(U)\ni [x]$ and $[x]\notin\pi(V)\ni [y]$.
If $C \ni x\neq y \in X-C$. I considered two open sets $A,B\subset X$ such that $C \subset A \ni x$ and $y \in B \subset X-C$ such that $x \notin B$ and $y \notin A$ (Hausdorff $\implies$ $T_1$). Note that $A$ is saturated since $\pi^{-1}(\pi(A))=A$, because all the fibres of classes that identify $C$ give $C$, and the elements $[z]$
in $\pi(A-C)$ give $z$, then $\pi(A)$ is open in the quotient topology. So, as I wrote in $(\square)$,  $X/\sim$ is $T_1$.
The part "$2)$" seems too similar to part "$1)$"... Am I missing something?
Thank you in advance.
 A: You’ve made the first part much harder than necessary. For convenience let $Y=X/\sim$, let $[x]\in Y$, and let $U=Y\setminus\{[x]\}$. If $x\in X\setminus C$, then $\pi^{-1}[U]=X\setminus\{x\}$, which is open in $X$, so $U$ is open in $Y$, and $\{[x]\}$ is closed. If $x\in C$, then $\pi^{-1}[U]=X\setminus C$, which is open in $X$, so $U$ is again open in $Y$, and $\{[x]\}$ is closed. Thus, each singleton $\{y\}$ in $Y$ is closed, so $Y$ is $T_1$. (If you’ve not already shown that a space is $T_1$ iff each singleton set in it is closed, this would be a good time to prove that useful fact.)
However, the extra effort isn’t really wasted, because the proof that $Y$ is Hausdorff is, as you suspect, very similar to what you’ve already done. If $x$ and $y$ are distinct points of $X\setminus C$, they have disjoint open nbhds disjoint from $C$, and $\pi$ maps those nbhds to disjoint open nbhds of $\pi(x)$ and $\pi(y)$. Regularity is needed only to separate $[x]$ and $[y]$ when one of $x$ and $y$ is in $C$.
