So im doing an exercise from Munkres, that is that if we have a perfect map $p : X \rightarrow Y$ then $X$ secound countable implies $Y$ secound countable. Im having some trouble seeing that the sets that we are going to consider are actually countable.
So my thought process was , well i have $y \in V$ open in $Y$, then $p^{-1}(y) \subset p^{-1}(V)$, well then since one is compact and the other one is open in $X$ and we have a basis we know that $p^{-1}(y) \subset \cup^{N}_{i=1}B_i$ where $B_i$ are basis elements of X. Then we can use the tube lemma and find $W$ neighborhood of $y$ such that $p^{-1}(W) \subset \cup^{N}_{i=1}B_i \subset p^{-1}(V)$ and then since the function $p$ is surjective $ W \subset Y$.
There is a hint in the exercise to consider the union of all sets $p^{-1}(W)$ sucht that there are countained in a finite union of elements of the basis of $X$, i just dont see how this is going to be a countable basis for $Y$, Thanks in advance.
Let $\mathcal{B} = \{B_n : n \in \mathbb{N}\}$ be a countable base for $X$.
By standard set theory, the set
$$\mathcal{B}':=\{O: \exists F \subseteq \mathbb{N} \text{ finite }: O= \bigcup_{n \in F} B_n\}$$
(all finite unions of subfamilies of $\mathcal{B}$) is also countable (essentially because the set of finite subsets of a countable set is still countable).
Now given $p$, define $W(O)=Y\setminus p[X\setminus O]$, which is open in $Y$ as $p$ is closed and obeys $p^{-1}[W(O)] \subseteq O$, and whenever $p^{-1}[\{y\}]\subseteq O$ we have $y \in W(O)$.
Then adapt your argument slightly to see that
$$\{W(O): O \in \mathcal{B}'\}$$ is a countable (because it's indexed by a countable set $\mathcal{B}'$) base for $Y$.