I came up with the following proof yet feel like it's wrong (justification for this feeling at the end):
Thorem. Let $(X,\mathscr{T}_X), (Y,\mathscr{T}(p))$ be spaces and $p\colon X\to Y$ an identification. Then $p$ is open.
Proof. Let $U$ be open in $X$. If $p(U)$ is open we are done. Since $p^{-1}(p(U))=U\in\mathscr{T}_X$, we have that the inverse image of $p(U)$ is open so, since the elements in $\mathscr{T}(p)$ are those and only those for which their inverse image through $p$ is open in $X$, $p(U)$ is then open $\square$.
I fill like something is wrong. I remember finding a criterion to know when an identification is open (which seems kinda pointless if what I just said is true). Is this proof correct?
It's not true that $p^{-1}(p(U)) = U$. For example, consider the quotient map $p:[0,1] \to S^1$ given by $p(t) = e^{2\pi i t}$. It isn't open since e.g. $U = [0,\frac12)$ is open but $p(U)$ isn't. And indeed $p^{-1}(p(U)) = U \cup \{1 \} \neq U$.