Prove that $A$ is closed. let $f:X\to Y$ continuous, open and onto. Then $ Y $ is Hausdorff if and only if the set $ \{(x, y): f (x) = f (y) \} $ is closed. I already did the one for $ Y$ is Hausdorff so the set is closed. Someone help me to come, I don't know how to do it.
Let $A=\{(x, y): f (x) = f (y) \}$. What I will prove is that the $ A'$ is open. Let $ x, y \in A'$. So, $ f (x) \neq f (y) $, since $ Y $ is $ T_2 $, then we have $ U_ {f (x)} $ and $ V_ {f (y)} $ two open such that $ U_ {f (x)} \cap V_ {f (y)} = \varnothing. $ Now, $$f^{-1}(U_ {f (x)}) \cap f^{-1}(V_ {f (y)})=f^{-1}(U_ {f (x)} \cap V_ {f (y)} )=f^{-1}(\varnothing)=\varnothing $$
From there we will already obtain that $ A '$ is open, so $ A $ is closed. The other implication is missing.

Suppose $A$ is closed and let $y_1 \neq y_2$ in $Y$. By ontoness there are $x_1, x_2 \in X$ so that $f(x_1)=y_1, f(x_2)= y_2$. By definition $(x_1,x_2) \in X^2\setminus A$ and as $A$ is closed, there is a basic open subset $U \times V$ of $X^2$ so that $$(x_1, x_2) \in U \times V \subseteq X^2\setminus A\tag{1}$$
Now if $z \in f[U] \cap f[V]$ there would be $x_3 \in U, x_4 \in V$ so that $z=f(x_3)=f(x_4)$ but then $(x_3,x_4) \in (U \times V)\cap A$ would contradict the inclusion in $(1)$. So $f[U] \cap f[V]=\emptyset$.
By openness of $f$, $f[U]$ and $f[V]$ are the required disjoint open neighbourhoods of $y_1, y_2$ resp. and $Y$ is $T_2$.
