If $f:X\rightarrow Y$ is a continuous open map then so is $f:X\rightarrow f(X)$
My attempt:
Let $U$ be open in $f(X)$ so $U=U' \cap f(X)$ where $U'$ is an open subset of $Y$. Then $f^{-1}(U' \cap f(X))$ $=$ $f^{-1}(U')$, which is open in $X$. Hence it is continuous.Let $U$ be open in $X$. So $f(U)$ is open in $Y$. Since $U\subseteq X$, it follows that $f(U)\subseteq f(X)$ and thus $f(U)=f(U)\cap f(X)$, which is open in $f(X)$.
Is my attempt correct?
Cleaner is to use a separate notation for the codomain-restricted $f$:
$$f': X \to f[X] \text{ defined by } \forall x \in X: f'(x)=f(x)$$
and $f[X]$ inherits the topology from $Y$.
Both parts of your (correct) proof can be shortened by knowing "open in open is open" and given that $f[X]$ is open by assumption we just say:
If $O \subseteq f[X]$ is open then $O$ is open in $Y$ and so $f'^{-1}[O]=f^{-1}[O]$ is open by continuity of $f$. So $f'$ is continuous.
If $O \subseteq X$ is open, $f'[O]=f[O]$ is open in $Y$ and a subset of $f[X]$ so open in $f[X]$ too. So $f'$ is open.
Two sentences, that's all it takes.