Covering space of a compact connected surface without boundary is a compact surface without boundary

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Let $p: X' \rightarrow X$ be an n-sheeted cover of $X$. I proved that $X$ being compact implies that $X'$ is compact in the standard way. I started with an open cover of $X'$, projected it under $p$, used compactification to get a finite subset that covers $X$, and then took the inverse image of those subsets to get a finite sub collection of the original open cover.

How can I use this local information in conjunction with compactness of $X'$ and $X$ to show that $X'$ does not have a boundary?

Also, how can I show that $E(X')=nE(X)$ where $E$ is the Euler characteristic.

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Let $x'\in X'$. Write $p(x')=x$. Let $U$ an open subset which is a chart of $X$ and which contains $x$ such that $p^{-1}(U)=V_1,...,V_n$ such that the restriction $p:V_i\rightarrow U$ is a diffeomorphism. Suppose $x'\in V_{i_0}$. Since $U$ is an open subset of $\mathbb{R}^2$ (the domain of a chart) so is $V_{i_0}$, we deduce that $x'$ has an open neighborhood diffeomorphic to an open subset of $\mathbb{R}^2$ and that the boundary of $X'$ is empty.