Ok, I've been struggling with the following problem; before I get into it I want to put a disclaimer that I only want a hint, not a solution.
Let $X$ be a topological space, consider $\sim$ defined by $x \sim y \iff \overline{\{ x \}} = \overline{\{ y \}}$, show that $X / \sim$ is $T_0$.
I've shown that $\sim$ is an equivalence relation, but am really struggling showing $X / \sim$ is $T_0$.
My start: In order to show $X$ is $T_0$ I aim to show that if $\tilde{x} \neq \tilde{y}$ then $\overline{\{\tilde{x}\}} \neq \overline{\{\tilde{y}\}}$ given $\tilde{x}, \tilde{y} \in X / \sim$. I've shown that $\overline{\tilde{x}} = \overline{\{x\}}$ for any $x \in \tilde{x}$ (where the closure is in the sense of the topology on $X$) so that $\overline{\tilde{x}}$ is disjoint from $\overline{\tilde{y}}$ for any distinct $\tilde{x}, \tilde{y}$. The issue I keep running into is relating $X$ to $X / \sim$, for example I'm not sure how to use the fact that $\overline{\tilde{x}}$ is disjoint from $\overline{\tilde{y}}$ to show $\overline{\{\tilde{x}\}} \neq \overline{\{\tilde{y}\}}$.
EDIT: I was pretty off I think, but I think I have a better handle on this now.
Put $q$ to be the natural map from $X \to X / \tilde{}$, and let $x,y\in X$ be such that $q(x) \neq q(y)$.
For any $x \in X$ we have $\operatorname{cl}_X \{ x \} = \operatorname{cl}_X q^{-1}\left[ \{ q(x) \} \right]$ since $$ x \in q(x) \implies x \in q^{-1}\left[ \{ q(x) \} \right] \implies x \in \operatorname{cl}_X q^{-1}\left[ \{ q(x) \} \right] \implies \operatorname{cl}_X \{x\} \subset \operatorname{cl}_X q^{-1}\left[ \{ q(x) \} \right] $$ and if $y \in q^{-1}\left[ \{ q(x) \} \right]$ then $\operatorname{cl}_X \{x\} = \operatorname{cl}_X \{y\}$ so $y \in \operatorname{cl}_X \{x\}$ thus $\operatorname{cl}_X q^{-1}\left[ \{ q(x) \} \right] \subset \operatorname{cl}_X \{ x \}$.
Next notice since $q(x) \neq q(y)$ we have $\operatorname{cl}_X \{ x \} \neq \operatorname{cl}_X \{ y \}$. W.l.o.g. consider $\operatorname{cl}_X \{ x \} \setminus \operatorname{cl}_X \{ y \} \neq \emptyset$ and note that $\exists z \in q^{-1}\left[ \{ q(x) \} \right] \setminus \operatorname{cl}_X \{ y \}$, otherwise if $\operatorname{cl}_X \{ y \} \supset q^{-1}\left[ \{ q(x) \} \right]$ then $\operatorname{cl}_X \{ y \} \supset \operatorname{cl}_X q^{-1}\left[ \{ q(x) \} \right]$, but we have $\operatorname{cl}_X \{ x \} = \operatorname{cl}_X q^{-1}\left[ \{ q(x) \} \right]$ so we would get $\operatorname{cl}_X \{y\} \supset \operatorname{cl}_X \{x\}$ which tells us $\operatorname{cl}_X \{x\} \setminus \operatorname{cl}_X \{y\} = \emptyset$, a contradiction to our assumption. Note also that $z \in q^{-1}\left[ \{ q(x) \} \right] \implies q(z) = q(x)$.
Now I want to show that $q \left[ X - \operatorname{cl}_X \{ y \} \right]$ is open, to do this I will show that $$ q^{-1} \left[ q \left[ X \setminus \operatorname{cl}_X \{ y \} \right] \right] = X \setminus \operatorname{cl}_X \{ y \} $$ First let $w \in X \setminus \operatorname{cl}_X \{y\}$ then $q(w) \in q \left[X \setminus \operatorname{cl}_X \{y\} \right]$ so $w \in q^{-1} \left[ q \left[X \setminus \operatorname{cl}_X \{y\}\right] \right]$. Next let $w \in q^{-1} \left[ q \left[X \setminus \operatorname{cl}_X \{y\} \right] \right]$ then $q(w) \in q \left[ X \setminus \operatorname{cl}_X \{ y \} \right]$ so that $\exists v \in X \setminus \operatorname{cl}_X \{ y \}$ such that $\operatorname{cl}_X \{ v \} = \operatorname{cl}_X \{ w \}$. Now suppose for contradiction that $w \in \operatorname{cl}_X \{ y \}$ then $\operatorname{cl}_X \{ w \} \subset \operatorname{cl}_X \{ y \}$ so $\operatorname{cl}_X \{ v \} \subset \operatorname{cl}_X \{ y \}$ thus $v \in \operatorname{cl}_X \{ y \}$ but we have that $v \in X \setminus \operatorname{cl}_X \{ y \}$ so we conclude that $w \in X \setminus \operatorname{cl}_X \{ y \}$.
Now $q(y) \not\in q\left[ X \setminus \operatorname{cl}_X \{ y \} \right]$, otherwise this would tell us that $\operatorname{cl}_X \{ y \} = \operatorname{cl}_X \{ v \}$ for some $v \in X \setminus \operatorname{cl}_X \{ y \}$ so that $v \in X \setminus \operatorname{cl}_X \{ v \}$, a contradiction. Also $q(x) \in q \left[ X \setminus \operatorname{cl}_X \{ y \} \right]$ since $z \in X \setminus \operatorname{cl}_X \{ y \}$ and $q(z) = q(x)$. Thus I found an open set containing $q(x)$ but not $q(y)$ so $X / \tilde{}$ is $T_0$.
This feels fairly jenky, though, so I'm unsure if I missed something and did something wrong or if I missed something and the proof is much simpler.
HINT: Let $q:X\to X/\!\sim$ be the quotient map. You know that if $q(x)\ne q(y)$, then $$\operatorname{cl}_X\{x\}\ne\operatorname{cl}_X\{y\}\;.$$ Without loss of generality there is a point $z\in(\operatorname{cl}_X\{x\})\setminus\operatorname{cl}_X\{y\}$; note that $q(z)=q(x)$. Show that $q^{-1}\left[q\big[\operatorname{cl}_X\{y\}\big]\right]=\operatorname{cl}_X\{y\}$, and consider the set $q\big[X\setminus\operatorname{cl}_X\{y\}\big]$ in $X/\!\sim$.
(If you’re familiar with the term saturated set in this context, I’m suggesting that you show that $\operatorname{cl}_X\{y\}$ is saturated and see what this tells you about its complement.)