Let $f: X\rightarrow Y$ be a continuous open map. Show that if $X$ satisfies the second countability axiom, then so does $f(X)$.
My attempt:Let $B_1,B_2,B_3....$ be an enumeration of the basis $\mathbb{B}$ the topology on $X$ admits. Then clearly, $A=$ $\{$ $f(B_i)$ $:$ $B_i \in \mathbb{B}$ $\}$ is countable. Further observe that for each $B\in \mathbb{B}$ , $f(B)=f(B)\cap f(X)$, and so $A$ is a collection of sets open relative to $f(X)$. Let $U$ be an open subset of $f(X)$. Then $f^{-1}(U)=f^{-1}(U' \cap f(X))$ ( $U'$ is open in $Y$ ) $=$ $f^{-1}(U')\cap X =f^{-1}(U')$, which is open in $X$ since $f$ is continuous. So $f^{-1}(U)=f^{-1}(U')$ is the union of some countable collection $(B_j)_{j \in J} \subseteq \mathbb{B}$. Hence $U=\bigcup_{j\in J}f(B_i)$. Since $U$ is arbitrary, $A$ is a countable basis for $f(X)$.
Is my attempt correct? (Please answer this question)
By the way, I so appreciate the help I have received in my previous posts, the feedback is soooooo helpful.
It can be slightly simplified:
Note that $f[X]$ is open in $Y$ and indeed if $\Bbb B=\{B_n: n \in \Bbb N\}$ is a countable base for $X$, the sets $f[\Bbb B]:=\{f[B_n]: n \in \Bbb N\}$ are open sets in $Y$ and thus open sets of $f[X]$ too.
And if $U \subseteq f[X]$ is open in $f[X]$, it's also open in $Y$ and so $f^{-1}[U]$ is open in $X$. So for some subset $J$ of $\Bbb N$ we have
$$f^{-1}[U]= \bigcup_{n \in J} B_n$$ and so noting that $f$ preserves unions and $f[f^{-1}[U]] = U$ we get
$$U = f[f^{-1}[U]] = f[\bigcup_{n \in J} B_n] = \bigcup_{n \in J} f[B_n]$$
and so $U$ is a union of sets from $f[\Bbb B]$ and so that latter collection is a base for $f[X]$.