Open surjective continuous functions being closed.

Let there be an open surjective continuous function $f:X\to Y$, where $X$ and $Y$ are topological spaces. I am given to understand that this mapping need not be closed. But could you point out the flaw in the following proof?

Let the open set $A\subseteq X$ map to the open set $f(A)$. Also assume $f$ is surjective. Then the complement of $A$ or $A'$ maps to $f(A)'$. Both are closed as per the definition of closed sets.

I know counter-examples exist. I'm just looking to find the flaw.

Thanks in advance!

The problem is that $X\setminus A$ need not map to $Y\setminus f[A]$. Consider the map

$$\pi:\Bbb R^2\to\Bbb R:\langle x,y\rangle\mapsto x$$

that projects the plane to the $x$-axis. This map is continuous and open. The set $L=\Bbb R\times\{0\}$ is closed, and $\pi[L]=\Bbb R$, but $\pi[\Bbb R^2\setminus L]=\Bbb R$ as well.

Because of $f(A^C)\neq f(A)^C$ just think of constant functions. Here $A^C$ is the complement of $A$.

When your function is surjective you only know that

$f(A^C)\supseteq f(A)^C$ and equality holds when $f$ is bijective.