$\def\RR{\mathbb{R}}$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 $V$.
I guess your question is whether your proof is correct although you didn't say this. Yes, it is.
Note that a function $p : X \to Y$ is a quotient map in the sense of your definition if and only if the following are satisfied:
(1) $p$ is surjective.
(2) A subset $U \subset Y$ is open in $Y$ if and only if $p^{-1}(U)$ is open in $X$.
This is the usual definition of a quotient map. The continuity of $p$ is nothing else than the "only if" part in (2). To verify the equivalence of both definitions observe that the saturated [open] subsets of $X$ are precisely the sets having the form $p^{-1}(B)$ [with $B \subset Y$ open]. This is true because $p(p^{-1}(B)) = B$ by surjectivity of $p$.
If you start with the above definition, then your problem is equivalent to showing that if $V \subset Y$ is open , then $q : p^{-1}(V) \to V$ satisfies the "if" part of (2). So let $U \subset V$ be a set such that $W = q^{-1}(U)$ is open in $p^{-1}(V)$. Since $p^{-1}(V)$ is open in $X$, also $W$ is open in $X$. But $W = p^{-1}(U)$, hence $U$ is open in $Y$. This implies that $U$ is open in $V$.
Best Answer
Yes, this is classical: if $f: X \to Y$ is a quotient map (in the sense that $U \subseteq Y$ open iff $f^{-1}[U]$ open in $X$). Then define $$\forall x,x' \in X: x \sim x' \iff f(x)=f(x')$$
It's trivial to check this is an equivalence relation on $X$, the equivalence relation induced by $f$ (I've seen $R_f$ used for this as well). I'll denote the class of $x$ by $[x]$, and the map $x \to [x]$ by $q: X \to X{/}\sim$.
Then the map $$h: (X{/}\sim) \to f[X]; h([x]) = f(x)$$
is well-defined and a homeomorphism between $X{/}\sim$ and $f[X]$, where of course $X{/}\sim$ has the quotient topology w.r.t. $q$, as expected, and $f[X]$ the subspace topology w.r.t. $Y$.
Sketch of proof: that $h$ is a bijection is set-theoretically obvious; that $h$ is continuous follows from $h^{-1}[U \cap f[X]] = q[f^{-1}[U]]$, which is open in $X{/}\sim$ (when $U \subseteq Y$ is), as the $q$-image of a $\sim$-saturated open set etc. and if $O \subseteq X{/}\sim$ is open, $h[O]=f[q^{-1}[O]]$ which is open in $Y$ (and $f[X]$) by the quotient-property of $f$.
In short, if $f$ is quotient and surjective (aka onto), indeed $Y=f[X]$ can be seen as an "equivalence-relation-quotient" of $X$ as well. This "validates" the name quotient map for maps like $f$, one could say.