Point-Set Topology: Separation Axiom Proof Let $X$ and $Y$ be spaces and $f: X \rightarrow Y$ be a continuous, open surjection. Prove that $Y$ is Hausdorff if and only if $\{(x,y) \in X \times Y : f(x) = y\}$ (= graph of $f$) is closed in $X \times Y$.
The closed graph theorem states that if $X$ is a topological space and $Y$ is a compact Hausdorff space, then the graph of $f$ is closed if and only if $f$ is continuous.
My ideas:
For the $\Rightarrow$ direction, suppose that $Y$ is Hausdorff. Then, if we show that $Y$ is compact, and we know that $f$ is continuous, we can use the closed graph theorem to conclude that the graph of $f$ is closed in $X \times Y$. 
For the $\Leftarrow$ direction, I am not sure how we can use $f$ is a continuous open surjection and its graph is closed to show that $Y$ is Hausdorff. 
 A: If $Y$ is Hausdorff, $\Delta_Y=\{(y,y)\mid y \in Y\}$ is closed in $Y \times Y$. 
Moreover, $f$ being continuous implies that $f'=f \times 1_Y: X \times Y \to Y \times Y$ is continuous, where $f'(x,y)=(f(x),y)$. So $Y$ Hausdorff implies $\Gamma(f)=\{(x,y)\mid f(x)=y\} = f'^{-1}[\Delta_Y]$ is closed.
Note that we don't use anything about $f$ except it being continuous.
We do need something about $f$ to get Hausdorffness on $Y$ (a constant map will have a closed graph if $Y$ is $T_1$, e.g.).
So assume that $\Gamma(f)$ is closed, and $y \neq y'$ are two points of $Y$ we need to separate. We can write $y=f(x), y'=f(x')$ for some $x,x' \in X$ by surjectivity of $f$.
Note that  $(x,y') \notin \Gamma(f)$ as $f(x)=y \neq y'$ so we have a basic open set $U_{x'} \times V_y$ such that
$$(x,y') \in U_{x'} \times V_y \subseteq (X \times Y) \setminus \Gamma(f)$$ 
and similarly a basic open set for $(x,y') \notin \Gamma(f)$ as well, so that
$$\exists U_x, V_{y'} \text{ open }: x \in U_x, y' \in V_{y'} \text{ and } (U_x \times V_{y'}) \cap \Gamma(f)=\emptyset$$
Now check that $f[U_x]\cap V_y$ and $f[U_{x'}] \cap V_{y'}$ are disjoint, open (as $f$ is open) neighbourhoods of $y$ resp. $y'$ in $Y$, showing $Y$ to be Hausdorff indeed.
