Let $p \colon X \rightarrow Y$ be a closed continuous surjective map such that $p^{-1} \big( \{ y \} \big)$ is compact for each $y \in Y$. (Such a map is called a perfect map.) (c) Show that if $X$ is locally compact, then so is $Y$.
My attempt: Let $y\in Y$. $p^{-1}(y)$ is compact. Since $X$ is locally compact, we have $\forall x\in p^{-1}(y)$, $\exists C_x\subseteq X$, $\exists U_x \in \mathcal{N}_x$ such that $U_x \subseteq C_x$ and $C_x$ is compact. So $\{U_x |x\in p^{-1}(y)\}$ is an open cover of $p^{-1}(y)$. Since $p^{-1}(y)$ is compact, $\exists \{U_{x_1},..,U_{x_n}\}$ finite subcover of $\{U_x |x\in p^{-1}(y)\}$. $p^{-1}(y)\subseteq \bigcup_{i=1}^n U_{x_i}=U$. By exercise 3 section 26, $C=\bigcup_{i=1}^n C_{x_i}$ is compact. $U\subseteq C$. So $p(U)\subseteq p(C)$. Since $p$ is continuous, $p(C)$ is compact. By generalize hint/chapter 3 theorem 11.2 of Dugundji topology, $\exists V\in \mathcal{N}_y$ such that $p^{-1}(V)\subseteq U$. Since $p$ is surjective, $V=p(p^{-1}(V))\subseteq p(U)$. Thus $V\subseteq p(U)\subseteq p(C)$. Hence $\exists p(C)\subseteq Y$, $\exists V\in \mathcal{N}_y$ such that $V \subseteq p(C)$ and $p(C)$ is compact. Is this proof correct?
Edit: Unlike exercise 7(a) and exercise 7(b), in this proof we used all conditions of perfect map.
