I am asked to prove or disprove that if $X$ is $T_0$ and $f: X \longrightarrow Y$ is continuous, open and surjective then $Y$ is also $T_0$.
At first I tried to find any counterexample but I didn't find any. So I tried to prove it.
My attempt was that if $y_1 , y_2 \in Y$ are different points then by $f$ being surjective $f^{-1}(\{y_1\}) , f^{-1}(\{y_2\})$ are both non empty. If there exists $x \in f^{-1}(\{y_1\})$ with a neighbourhood $V$ of $x$ that has no points in common with $f^{-1}(\{y_2\})$ then $f(V)$ is a neighbourhood that contains $y_1$ but not $y_2$. Analogously, if there exists $x \in f^{-1}(\{y_2\})$ with a neighbourhood that contains no points of $f^{-1}(\{y_1\})$ then the same holds.
So if those points doesn't exist, that means that $f^{-1}(\{y_2\})\subseteq \overline{f^{-1}(\{y_1\})}$ and $f^{-1}(\{y_1\}) \subseteq \overline{f^{-1}(\{y_2\})}$. I think this may contradict that $X$ is $T_0$ but I am struggling to prove it (if it's true). Any help? Thanks in advance.
You're struggling because the claim is not true. An elementary example is as follows:
let $X=(\mathbb{Z};\text{cofinite})$ (that is, $S\subseteq X$ is open iff $S$ is empty or $X\setminus S$ is finite) and $Y=(\mathbb{Z}/(2);\text{indiscrete})$. Choose $f$ to be the quotient map (in the algebraic sense) which sends every $n$ to its remainder modulo $2$.
As a quotient of rings, $f$ must be surjective; since the indiscrete topology is terminal, $f$ must be continuous; and the image of every nonempty open set is all of $\mathbb{Z}/(2)$, so that $f$ is open. (All of these are also easy to check "by hand".)
But $X$ is $T_0$ and $Y$ is not.