Problem: Let $f:X\rightarrow Y$ be continuous and open. Suppose $X$ satisfies first countability axiom. Show that $f(X)$ satisfies first countability axiom.
My attempt: Let $b\in f(X)$ So there is an a $\in X$ such that $f(a)=b$. Let $U$ be an open subset of $f(X)$ containing $b$. So $U = U_b' \cap f(X)$. where $U_b'$ is open in $Y$. Since $X$ is open in $X$, $X=\bigcup_{p\in X,B\in \mathbb{B_p}}B$ where $\mathbb{B_p}$ is a neighborhood basis. So $U=\bigcup_{p\in X,B\in \mathbb{B_p}}(U_b' \cap f(B))$ which is open in $f(X)$. Hence $\{$ $U_b' \cap f(B)$ $:$ $b\in f(X)$, $B\in \mathbb{B_p}, p\in X$ $\}$ is a countable basis for $f(X)$ at $b$.
Is my proof correct?
Make it better structured:
Suppose $y \in f[X]$ and let $x \in X$ be such that $f(x)=y$. It suffices to show $y$ has a countable local base.
Let $\mathcal{B}_x$ be a countable local neighbourhood base for $x$ ($X$ is first countable). I claim that
$$\mathcal{B}'_y:= \{f[B]: B \in \mathcal{B}_x\}$$ is a (countable) local base at $y$.
Proof: First note that the sets in $\mathcal{B}'_y$ are open by openness of $f$ (an assumption you mentioned in the title but not in the proof!) and all contain $y$. So that out of the way, let $O$ be open in $f[X]$ and $y \in O$. As again $f[X]$ is open in $Y$ we also know that $O$ is open in $Y$. And so $x \in f^{-1}[O]$ and the latter set is open by continuity of $f$, there is some $B \in \mathcal{B}_x$ such that $B \subseteq f^{-1}[O]$. Hence $f[B] \in \mathcal{B}'_y$ and $f[B] \subseteq O$ and so $\mathcal{B}'_y$ is a local base at $y$. Finally, as $\mathcal{B}_x$ is countable, so is $\mathcal{B}'_y$ as its image under a surjective map.
The weird unions in the OP's proof are unnecessary. Just straightforwardly define a local base in $f[X]$ from one in $X$.