Let $p : X \longrightarrow Y$ be a covering map of finite degree with $Y$ compact. Then given $y \in Y$ and given an open neighborhood $U \supseteq p^{-1} (y)$ in $X$ there exists an open neighborhood $V_y \ni y$ in $Y$ such that $p^{-1} (V_y) \subseteq U.$
Could anyone give me some suggestion as to how to show that?
This is true for any covering map of finite degree $n$, we do not need the assumption that $Y$ is compact.
Let $W_y$ be an evenly covered open neighborhood of $y$. Then $p^{-1}(W_y) = \bigcup_{i=1}^n U_{y,i}$ with open $U_{y,i} \subset X$ such that the restrictions $p_{y,i} : U_{y,i} \xrightarrow{p} W_y$ are homeomorphisms. Note that for each $M \subset W_y$ we have $$p^{-1}(M) = p^{-1}(M) \cap p^{-1}(U) = p^{-1}(M) \cap \left(\bigcup_{i=1}^n U_{y,i}\right) = \bigcup_{i=1}^n(p^{-1}(M) \cap U_{y,i}) \\ = \bigcup_{i=1}^n p_{y,i}^{-1}(M) .$$
Given an open neighborhood $U$ of $p^{-1}(y)$, let $V_{y,i} = p(U \cap U_{y,i}) = p_{y,i}(U \cap U_{y,i})$. The $V_{y,i}$ are open subsets of $Y$. Define $$V_y = \bigcap_{i=1}^n V_{y,i} .$$ This is an open subset of $Y$ contained in $W_y$. We have $$p^{-1}(V_y) = \bigcup_{i=1}^n p_{y,i}^{-1}(V_y) \subset \bigcup_{i=1}^n p_{y,i}^{-1}(V_{y,i}) = \bigcup_{i=1}^n (U \cap U_{y,i}) \subset U. $$