$\def\RR{\mathbb{R}}$There is no separation condition which will do the job. That's a vague statement, so here is a precise one: There is a subset of $\mathbb{R}^2$ which (equipped with the subspace topology) does not have condition $\dagger$.
Proof: Let $A$ and $B$ be two disjoint dense subsets of $\mathbb{R}$, neither of which contains $0$. (For example, $\mathbb{Q} +\sqrt{2}$ and $\mathbb{Q}+\sqrt{3}$.) Let
$$X = (A \times \RR_{\geq 0}) \cup (B \times \RR_{\leq 0}) \cup (\{0\} \times \RR_{\neq 0}) \subset \RR^2.$$
Define $(x_1,y_1)$ and $(x_2, y_2)$ to be equivalent if $x_1=x_2$ and, in the case that $x_1=x_2=0$, that $y_1$ and $y_2$ have the same sign.
Verification that this is a closed equivalence relation: $X^2$ is a metric space, so we can check closure on sequence. Let suppose we have a sequence $(x_n, y_n) \sim (x'_n, y'_n)$ with $\lim_{n \to \infty} x_n=x$, $\lim_{n \to \infty} y_n=y$, $\lim_{n \to \infty} x'_n=x'$ and $\lim_{n \to \infty} y'_n=y'$. We must verify that $(x,y) \sim (x',y')$. First of all, we have $x_n = x'_n$, so $x=x'$ and, if $x=x' \neq 0$, we are done. If $x=x'=0$, we must verify that $y$ and $y'$ have the same sign. But $y_n$ and $y'_n$ weakly have the same sign for all $n$, so they can't approach limits with different signs.
Verification that $X/{\sim}$ is not Hausdorff: We claim that no pairs of open sets in $X/{\sim}$ separates the images of $(0,1)$ and $(0,-1)$. Suppose such open sets exist, and let $U$ and $V$ be their preimages in $X$. Then there is some $\delta$ such that $(A \cap (-\delta, \delta) )\times \RR_{\geq 0} \subset U$ and $(B \cap (-\delta, \delta) )\times \RR_{\geq 0} \subset B$. Then $U \cap \RR \times \{ 0 \}$ is an open set which contains $(-\delta, \delta)$. By the density of $B$, there must be a point of $B \cap (- \delta, \delta)$ in $U \cap \RR$, and then this gives an intersection between $U$ and $V$.
The answer is "yes", and the result can be found in Bourbaki's General Topology, Part I, Exercise 19 to $\S$ 10 of Chapter 1,
on page 153 of the English translation published by Springer.
The exercise has a hint, but it took me quite a time to fully elaborate the proof.
So I try to present the argument with some more details included.
Let us first make clear some terminology.
A neighbourhood of a set $A$ is any set containing an open set containing $A$.
A space $X$ is compact if any open cover of $X$ has a finite subcover.
A space is locally compact if it is Hausdorff and any point has a compact neighbourhood;
it is $\sigma$-compact if it is a countable union of compact sets.
A $\sigma$-locally compact space is a space which is both $\sigma$-compact and locally compact.
A normal space is a space in which any two disjoint closed sets have disjoint neighbourhoods.
It is well known that every compact Hausdorff space is normal, and
in a normal space, any two disjoint closed sets have disjoint closed neighbourhoods.
A normal space in which every singleton is closed, is necessarily Hausdorff.
If $X$ is compact and $Y$ is arbitrary then the projection to the second coordinate $\textit{pr}_2\colon X\times Y\to Y$ is a closed mapping.
It is also known that every $\sigma$-locally compact Hausdorff space $X$ can be expressed as $\bigcup_{n\in\omega}X_n$,
where $X_n$ are open, $\overline{X_n}$ are compact, and $\overline{X_n}\subseteq X_{n+1}$, for every $n$.
Let $E$ be an equivalence relation on $X$.
Denote by $[A]$ the saturation of a set $A\subseteq X$, that is, $[A]=\{y\in X\!:(\exists x\in A)\,(x,y)\in E\}$.
If $Y\subseteq X$ then $E_Y=E\cap(Y\times Y)$ is an equivalence relation on $Y$.
The saturation of a set $A\subseteq Y$ with respect to $E_Y$ is $[A]\cap Y$.
If $A,B\subseteq Y$ are disjoint sets saturated with respect to $E_Y$ then their saturations $[A]$, $[B]$ with respect to $E$ are disjoint as well.
Lemma.
Let $X$ be a compact Hausdorff space and let $E$ be an equivalence relation on $X$ such that $E$ is a closed subset of $X\times X$.
Then the quotient mapping $q\colon x\mapsto[x]$ is closed.
Proof.
Let $A$ be a closed subset of $X$.
Then $A$ is compact and $[A]=\textit{pr}_2(E\cap(A\times X))$.
Since $E\cap(A\times X)$ is closed in $A\times X$ and $\textit{pr}_2$ is a closed mapping, $[A]$ is a closed subset of $X$.
Hence $q$ is closed.
q.e.d.
Lemma.
Let $X$ be a normal space and let $E$ be an equivalence relation on $X$ such that the quotient mapping $q\colon x\mapsto [x]$ is closed.
Then the quotient space $X/E$ is normal.
Proof.
Let $A,B$ are disjoint closed saturated subsets of $X$.
Since $X$ is normal, there exist disjoint open sets $U,V$ such that $A\subseteq U$ and $B\subseteq V$.
Then $[X\setminus U]$ and $[X\setminus V]$ are closed saturated sets disjoint from $A$ and $B$, respectively,
hence $U'=X\setminus [X\setminus U]$ and $V'=X\setminus [X\setminus V]$ are open saturated neighbourhoods of $A$ and $B$,
respectively.
If $z\in U'\cap V'$ then $[z]$ is disjoint from both $X\setminus U$ and $X\setminus V$, hence $z\in U\cap V$, a contradiction.
It follows that $U',V'$ are disjoint,
so any two disjoint closed subsets of $X/R$ have disjoint open neighbourhoods.
q.e.d.
Theorem.
Let $X$ be a $\sigma$-locally compact Hausdorff space and let $E$ be an equivalence relation on $X$ such that $E$ is a closed
subset of $X\times X$. Then the quotient space $X/E$ is normal and Hausdorff.
Proof.
Let $X=\bigcup_{n\in\omega}X_n$, where $X_n$ are open sets such that $\overline{X_n}$ are compact and
$\overline{X_n}\subseteq X_{n+1}$ for every $n$.
For every $n$ let us consider the quotient space $Y_n=\overline{X_n}/E_n$, where $E_n=E\cap(\overline{X_n}\times\overline{X_n})$.
Since $\overline{X_n}$ is a compact Hausdorff space, the corresponding quotient mapping $q_n\colon\overline{X_n}\to Y_n$ is closed and hence the space $Y_n$ is normal.
Let $A,B$ be disjoint closed saturated subsets of $X$.
We prove that there exist disjoint open saturated sets $U,V$ such that $A\subseteq U$ and $B\subseteq V$.
For $n\in\omega$, define sets $A_n$, $B_n$, $U_n$, $V_n$ by induction as follows.
Start with $A_0=A\cap\overline{X_0}$, $B_0=B\cap\overline{X_0}$.
In $n$-th step, assume that $A_n$, $B_n$ are disjoint closed subsets of $\overline{X_n}$ saturated with respect to $E_n$.
Assume also that $A\cap\overline{X_n}\subseteq A_n$ and $B\cap\overline{X_n}\subseteq B_n$.
Then, by normality of $Y_n$, there exist disjoint relatively open subsets $U_n$, $V_n$ of $\overline{X_n}$ such that $A_n\subseteq U_n$, $B_n\subseteq V_n$, $\overline{U_n}\cap\overline{V_n}=\emptyset$, and $U_n$, $V_n$, $\overline{U_n}$, $\overline{V_n}$ are saturated with respect to $E_n$.
We have $\overline{V_n}\subseteq\overline{X_n}$, hence $$A\cap\overline{V_n}\subseteq A\cap\overline{X_n}\cap\overline{V_n}\subseteq A_n\cap\overline{V_n}\subseteq U_n\cap\overline{V_n}=\emptyset,$$ similarly $B\cap\overline{U_n}=\emptyset$.
Since $A,B$ are saturated with respect to $E$, we also have $A\cap\big[\,\overline{V_n}\,\big]=B\cap\big[\,\overline{U_n}\,\big]=\emptyset$.
Let us take $A_{n+1}=\big(A\cup\big[\,\overline{U_n}\,\big]\big)\cap\overline{X_{n+1}}$, $B_{n+1}=\big(B\cup\big[\,\overline{V_n}\,\big]\big)\cap\overline{X_{n+1}}$.
It is easy to check that $(A\cap\overline{X_{n+1}})\cup U_n\subseteq A_{n+1}$, $(B\cap\overline{X_{n+1}})\cup V_n\subseteq B_{n+1}$, and that $A_{n+1}$, $B_{n+1}$ are closed and saturated with respect to $E_n$.
Since $\overline{U_n}$, $\overline{V_n}$ are disjoint subsets of $\overline{X_n}$ saturated with respect to $E_n$, their saturations $\big[\,\overline{U_n}\,\big]$, $\big[\,\overline{V_n}\,\big]$ with respect to $E$ are disjoint as well, and we obtain that $A_{n+1}\cap B_{n+1}=\emptyset$.
Hence we can proceed with the induction.
This way we can define increasing sequences $\{U_n\}_{n\in\omega}$, $\{V_n\}_{n\in\omega}$ such that $U_n$, $V_n$ are relatively open subsets of $\overline{X_n}$, $A\cap\overline{X_n}\subseteq U_n$, $B\cap\overline{X_n}\subseteq V_n$, $\overline{U_n}\cap\overline{V_n}=\emptyset$, and $U_n$, $V_n$, $\overline{U_n}$, $\overline{V_n}$ are saturated with respect to $E_n$.
Set $U=\bigcup_{n\in\omega}U_n$, $V=\bigcup_{n\in\omega}V_n$.
If $x\in U$ then there exists $n$ such that $x\in U_n\cap X_n\subseteq U$, where $U_n\cap X_n$ is open, hence $U$ is open.
Also, if $x\in[U]$ then there exists $n$ such that $x\in\big[\,\overline{U_n}\,\big]\cap\overline{X_{n+1}}$, hence $x\in U$.
So $U$ is an open set saturated with respect to $E$, similarly for $V$.
Finally, $U$ and $V$ are disjoint since if $x\in U\cap V$ then there exists $n$ such that $x\in U_n\cap V_n$, which is impossible.
We have proved that the quotient space $X/E$ is normal.
To see that it is also Hausdorff it suffices to show that every singleton in $X/E$ is a closed set.
This is clear since if $E$ is a closed subset of $X\times X$ then every equivalence class is closed.
q.e.d.
Best Answer
This would work if $X$ is, additionally, regular. Seperating the point $A$ from $\{x\}$ in $X/A$ is essentially the same as seperating the closed subset $A$ from the point $x$ in $X$, but this is not implied by the Hausdorff property - it is called regularity.
To understand this, let's think about what an open neighborhood of $A$ looks like in $X/A$. By definition, a neighborhood $A \in U \subseteq X/A$ is open iff $p^{-1}[U]$ is open in $X$. The preimage of $\{A\}$ is $A$, and the preimage of any other point $\{x\}$ is just $\{x\}$. So taking the preimage of our neighborhood $U$ has the effect of replacing the point $A \in X/A$ with the subset $A \subseteq X$. Thus, open neighborhoods of $A$ in $X/A$ are precisely those subsets which pull back to open subsets of $X$ containing $A$.
Now, suppose indeed that we could find disjoint open neighborhoods $A \in U$ and $\{x\} \in V$. Then we'd have $A \subseteq p^{-1}[U]$ and $x \in p^{-1}[V]$. But then these preimages are disjoint open subsets of $X$ which separate the subset $A$ from the point $x$. Thus, if the proposition you wrote was true then $X$ would have to be regular. Unfortunately, not all Hausdorff spaces are regular. See here for an example.