Finitely sheeted covering space of a compact space is compact Let $p:X\to Y$ be a finitely sheeted covering space. I want to show that $X$ is compact if $Y$ is. 
I have proven the following lemma.
Let $U\subset X$ be open containing $p^{-1}(b)$, then there exists an open subset $O_b$ such that $p^{-1}(O_b)\subset U$.
Now, let $\mathcal{U}$ be an open cover of $X$. I choose a finite evenly covered cover $\mathcal{V}$ of $Y$, which is possible since $Y$ is compact. Let $y\in Y$, then there exists an $U_i\in\mathcal{U}$ such that $y\in p(U_i)$. How can I go on and invoke the lemma?
 A: Let $\mathcal{U}$ be an open cover of $X$. For each $y \in Y$ the finite set $p^{-1}[\{y\}]$ is covered by finitely many members of $\mathcal{U}$, let's call these $\mathcal{U}_y \subseteq \mathcal{U}$ and let $O_y$ be open in $Y$ such that $p^{-1}[O_y] \subseteq \bigcup \mathcal{U}_y$, by your lemma. 
Then $\{O_y: y \in Y\}$ is an open cover of $Y$ so by compactness of $Y$ there are finitely many $O_{y_1}, \ldots, O_{y_n}$ that cover $Y$. Now note that $\bigcup_{i=1}^n \mathcal{U}_{y_i}$ is a finite (finite union of finite sets) subcover of $\mathcal{U}$:
$$X= p^{-1}[Y]=p^{-1}[\bigcup_{i=1}^n O_{y_i}] = \bigcup_{i=1}^n p^{-1}[O_{y_i}] \subseteq \bigcup_{i=1}^n \bigcup \mathcal{U}_{y_i}$$ 
Note that we only really use the property of the lemma (which is equivalent to $p$ being a closed map) and all fibres being compact, i.e. $p$ is a perfect map, and preimages of a compact space under a perfect map are compact, as this proof essentially shows.
A: Let $\mathcal U$ be an open cover of $X.$ Let $\mathcal V$ be the family of all open sets $V\subset Y$ such that there exists a finite subset $\mathcal U_V\subset\mathcal U$ with $p^{-1}(V)\subset \bigcup_{U\in \mathcal U_V}U.$ Use your lemma to show that $\mathcal V$ is a cover of $Y.$ A finite subcover $\mathcal V'\subset\mathcal V$ of $Y$ gives a finite subcover $\bigcup_{V\in\mathcal V'}\mathcal U_{V}\subset \mathcal U$ of $X.$
