$T_0$-identification of a topological space, proof using singleton closures In Willard's General Topology, section 13.2.c,
for any topological space X is defined a quotient space X/~ such that x ~ y iff $cl({\{x\}}) = cl(\{y\})$ where $cl(.)$ is the topological closure.
Then it is stated that if in the new space X/~ there are x and y such that their closure is the same, then they are actually equal.
Somewhere on the web (Jianfei Shen's "A Solution Manual for Willard (2004)" p38), I found this "proof" but this seems too much incomplete and does not satisfy me:

take any $cl(\{x\}) \neq cl(\{y\})$ in X/~. Then $cl(cl(\{x\})) = cl(\{x\}) \neq cl(\{y\}) = cl(cl(\{y\}))$

This is not clear at all why we should have $cl_{X/\sim}(cl_{X}(\{x\})) = cl_X(\{x\})$ as, inside $X/\sim$, the $cl_X(\{x\})$ are just standard elements (not sets, let alone closed sets) defined as equivalence classes from the relation $\sim$.
I've also looked into the quotient topology of $X/\sim$, where closed sets are the sets $C$ such that $\cup\{f^{-1}(y), y \in C\}$ is closed in $X$, where $f : x \in X \mapsto cl(\{x\}) \in X/\sim$
but I did not manage to sort out the path to the proof.
It should be obvious, from the fact that the author did not give any detailed explanation, but I am not getting it...
 A: Let $[x]$ denote the equivalence class of $x \in X$ with respect to $\sim$. We have $[x] \subset cl_{X}(\{x\})$ because if $y \sim x$, the $y \in cl_{X}(\{y\}) = cl_{X}(\{x\})$. However, in general $[x] \subsetneqq cl_{X}(\{x\})$. As an eaxmple take the Sierpinski space $X = \{0,1\}$ with open sets $\emptyset, \{0\}, X$. Then $cl_{X}(\{0\}) = X, cl_{X}(\{1\}) = \{1\}$. Hence
$[0] = \{0\} \subsetneqq cl_{X}(\{0\})$.
This shows that Jianfei Shen's solution manual erroneously assumes that the points of $X/\sim$ are the sets $cl_{X}(\{x\})$ with $x \in X$.
The space $X/\sim$ is nothing else than the Kolmogoroff quotient of $X$. In fact, we have $cl_{X}(\{x\}) = cl_{X}(\{y\})$ iff each open $U \subset X$ contains either both $x,y$ or none of $x,y$.
Now see Prove that the Kolmogorov Quotient is $T_0$ for a correct proof.
A: Answer based on this post from Paul Frost, and adapted to closed sets instead of open sets.
Firstly, let's prove that $\forall C \mathrm{closed} \subset X, f^{-1}(f(C)) = C$:
Let $a \in C \mathrm{closed} \subset X$, then $\forall b, (b \sim a) \Leftrightarrow (cl_X(\{a\})=cl(\{b\})) \implies (b \in C)$ because then all closed sets containing $a$ also contain $b$. So $[a] \subset C$.
Let $C$ a closed of $X$. Let $x \in f^{-1}(f(C))$. So $f(x) \in f(C)$. Choose $a \in C$ such that $f(x)=f(a)$.
So $x \sim a$ so $x \in [a] \subset C$. So $f^{-1}(f(C)) \subset C$. 
Obviously also $C \subset f^{-1}(f(C))$, so there is equality.
Secondly, $f$ is closed:
This is because $X_{/\sim}$ has by definition the quotient topology induced by $f$, so that the closed sets are the sets whose preimage by $f$ are closed in $X$.
So, since $\forall C \mathrm{closed} \subset X, f^{-1}(f(C)) = C$,
then $\forall C \mathrm{closed} \subset X, f(C) \mathrm{closed} \subset X_{/\sim}$.
Now let's prove the $T_0$-separability of $X_{/\sim}$:
$\forall x,y \in X_{/\sim} \mid x \neq y$,
$\forall x',y' \in X \mid (f(x')=x) \wedge (f(y')=y)$,
we have $cl_X(\{x'\}) \neq cl_X(\{y'\}))$
so that w.l.o.g. $\exists C \mathrm{closed} \subset X, (x' \in C) \wedge (y' \notin C)$.
Then $f(C) \mathrm{closed} \subset X_{/\sim}$ because $f$ is closed,
and $x \in f(C)$ but $y \notin f(C)$ because $f^{-1}(f(C))=C$ so that would mean $y' \in C$.
So $cl_{X/\sim}(\{x\}) \neq cl_{X/\sim}(\{y\})$.
