I want to show:
Let $f:X \to Z$ be a quotient map. Then $Z$ is Hausdorff if and only if whenever $x,y\in X$ and $f(x)\ne f(y)$, then there are $U,V$ neighborhoods of $x$ and $y$ respectively such that $U,V$ are disjoint and saturated with respect to $f$, i.e. $U=f^{-1}(f(U))$ and the same for $V$.
I have a problem with the implication $\Leftarrow$
Here is my argument:
I want to show that $Z$ is Hausdorff. Let $z\ne t\in Z$. Since $f$ is surjective, we have $t=f(x)$ and $z=f(y)$, and by hypothesis then there are $U,V$ neighborhoods of $x$ and $y$ respectively such that $U,V$ are disjoint and saturated with respect to $f$. Then we have $U=f^{-1}(U')$ and $V=f^{-1}(V')$ for some $U',V'$ subsets of $Y$. Is easily follows that $U'$ and $V'$ are disjoint and $t \in U'$ and $z \in V'$.
Problem how do I show that $U'$ and $V'$ are neighborhoods of $t$ and $z$ respectively?
For example, I have that there is an open set $A$ of $X$ with $x\in A \subseteq U$, but I don't know if $f(A)$ is open or not in $Y$. So how do I find an open subset of $Y$ wich is contained in $U'$ and contains $t$?
Warning: The definition of neighborhood we use here is : $U$ is a neighborhood of $x$ in $X$ if there is an open subset $A$ of $X$ such that $x \in A \subseteq U$. So here $U$ need not to be open. (See the comments to the answers below)
Since quotient maps transform saturated open subsets of the domain in open subsets of the codomain, I should find a saturated open subset $A\subseteq X$ such that $x\in A \subseteq U$. In this way $f(A)$ is open and $t\in f(A) \subseteq U'$. Any idea on how to do this?
If I understand it right then $U'=f(U)$. As you wrote you know that $f(U)$ contains the point $t$. Now why is it open? $f$ is a quotient map, so $f(U)$ is open in $Z$ if and only if its inverse image is open in $X$. But you know that $U=f^{-1}(f(U))$ and $U$ is an open set in $X$. Hence $U'=f(U)$ is open in $Z$, so it is a neighborhood of the point $t$. Now do the same thing for $V'$.