Exercise 12, Section 30 of Munkres’ Topology 
Let $f \colon X \rightarrow Y$ be a continuous open map. Show that if $X$ satisfies the first or the second countability axiom, then $f(X)$ satisfies the same axiom.

My attempt: Let $y\in f(X)$. Then $\exists x\in X$ such that $f(x)=y$. Since $X$ is first countable, $\exists \{U_n \in \mathcal{N}_x|n\in \Bbb{N}\}$ with the following property: $\forall U\in \mathcal{N}_x$, $\exists p\in \Bbb{N}$ such that $U_p \subseteq U$. Claim: $\{f(U_n)\in \mathcal{N}_y|n\in \Bbb{N}\}$ have the following property: $\forall V\in \mathcal{N}_y$, $\exists q\in \Bbb{N}$ such that $f(U_q)\subseteq V$. Proof: Since $U_n \in \mathcal{T}_X$, $\forall n\in \Bbb{N}$ and $f$ is open, we have $f(U_n)\in \mathcal{T}_Y$, $\forall n\in \Bbb{N}$. And $x\in U_n$, $\forall n\in\Bbb{N}$. So $f(x)=y \in f(U_n)$, $\forall n\in \Bbb{N}$. Thus $f(U_n)\in \mathcal{N}_y$, $\forall n\in \Bbb{N}$. let $V\in \mathcal{N}_y$. Then $f(x)\in V\in \mathcal{T}_Y$. Since $f$ is continuous, $f^{-1}(V) \in \mathcal{N}_x$. So $\exists m\in \Bbb{N}$ such that $U_m\subseteq f^{-1}(V)$. By elementary set theory, $f(U_m)\subseteq f(f^{-1}(V))\subseteq V$. Thus $\exists m\in \Bbb{N}$ such that $f(U_m)\subseteq V$. $f(X)$ have countable basis at $y$. Hence $f(X)$ is first countable.
Let $\mathcal{B}_X=\{B_n|n\in \Bbb{N}\}$ be a countable basis of $\mathcal{T}_X$. Claim: $\mathcal{B}’=\{ f(B_n)|n\in \Bbb{N}\}$ is countable basis of $\mathcal{T}_{f(X)}$. Proof: (1) Since map $f$ is open, $f(B_n)\in \mathcal{T}_Y$, $\forall n\in \Bbb{N}$. So $f(B_n)\cap f(X)=f(B_n)\in \mathcal{T}_{f(X)}$, $\forall n\in \Bbb{N}$. Thus $\mathcal{B}’\subseteq \mathcal{T}_{f(X)}$. (2) Let $V\in \mathcal{T}_{f(X)}$ and $y\in V$. By definition of subspace topology, $V=f(X) \cap U$; $U\in \mathcal{T}_Y$. $y\in f(X)$ and $y\in U$. So $\exists x\in X$ such that $f(x)=y\in U$. $x\in f^{-1}(U)$. Since $x\in f^{-1}(U)\in \mathcal{T}_X$, $\exists m\in \Bbb{N}$ such that $x\in B_m \subseteq f^{-1}(U)$. So $f(x)=y \in f(B_m)$ and $f(B_m)\subseteq f(f^{-1}(U)) \subseteq U$. Which implies $f(B_m)=f(B_m) \cap f(X) \subseteq U\cap f(X)=V$. Hence $\exists f(B_m)\in \mathcal{B}’$ such that $y\in f(B_m)\subseteq V$. By lemma 13.2, $\mathcal{B}’$ is basis of $\mathcal{T}_{f(X)}$. Is this proof correct?
 A: Both parts of your proof are correct.
However, since you included a proof-writing tag in your question, I'd like to add a quick comment about style! You use a lot of symbols, and that makes your proof a little bit difficult to read; you've also explicated some details that are already given in the premise (e.g., that $f(U) \in \mathcal{T}_Y$ for open $U$ in $X$; this is a restatement of the definition of an open map and doesn't require proof). Your proof might be shorter and a little easier to check if it's written in words and sentences, e.g.

Suppose $y \in f(X)$ and $V \subset f(X)$ is a neighborhood of $y$; then there exists $x \in X$ with $f(x) = y$. Since $f$ is continuous, $U = f^{-1}(V)$ is open. Because $X$ is first-countable, there is a countable local base $\mathcal{B}_x = \{U_n\}_{n=1}^\infty$ of $\mathcal{T}_X$ at $x$, giving us $U_k \subset U$ for some $k$. Since $f$ is an open map, $\mathcal{B}_y = \{f(U_n)\}_{n=1}^\infty$ is a collection of open neighborhoods of $y$. But $f(U_k) \subset V$, so $\mathcal{B}_y$ is a local basis of $\mathcal{T}_Y$ at $y$.

That said, not everyone likes paragraph proofs -- if you'd like to stick to using symbols and an iterated "Claim: Proof." structure, that's ok too! But I would then recommend taking special care to format with line breaks, bullet points, and similar presentational choices to make your argument easier to follow.
