Proposition. Let $\fE\stackrel{q}{\to}\cE\stackrel{p}{\to}X$ be covering spaces. Suppose that for every $x\in X$, $|p^{-1}(x)|<\infty.$ Then $p\circ q$ is a covering map.
Proof.
Let $U$ be an open neighborhood of a point $x$ such that
$$p^{-1}(U)=\bigsqcup_{i=1}^n \cV_i$$
where $\{\cV_i\}_{i=1}^n$ is a family of open sets in $\cE$, each of them mapped homeomorphically by $p$ onto $U$.
For each $i\in\{1,\dots,n\}$ take the unique element $e_i\in \cV_i$ such that $p(e_i)=x$.
For each $i$ take an evenly covered neighborhood $\cU_i$ of $e_i$ such that $\cU_i\subseteq \cV_i$ (If not, take the intersection). Take
$$U'=\bigcap_{i=1}^n p(\cU_i).$$
Then $U'$ is an open set containing $x$ (This, since each $\cU_i$ has a preimage of $x$).
Also, since $p_i=p|_{\cV_i}$ is a homeomorphism, then for each $i$, $\cU_i'=p_i^{-1}(U')$ is an open subset of $\cU_i$ mapped homeomorphically by $p$ onto $U'$. Also note that fixing $i$,
$$q^{-1}(\cU_i')=\bigsqcup_{j\in J}\fV_j$$
such that for each $j$, $q|_{\fV_j}$ is a homeomorphism onto $\cU_i'$. But then, since $p:\cU_i'\to U'$ is a homeomorphism, then $p\circ q:\fV_j\to U'$ is also a homeomorphism. And last,
$$p\circ q^{-1}(U')=\bigsqcup_{i=1}^n U_i'=\bigsqcup_{j_i\in J_i}\bigsqcup_{i=1}^n \fV_{j_i},$$
from which the result follows.
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