The disjoint union of covering spaces is again a covering space

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Let $B$ be an arc-connected and locally arc-connected space. Suppose that $p:E\to B$ and $q: F\to B$ are covering spaces. Let $r:E\sqcup F\to B$ be the function such that $r(x)=p(x)$ for all $x\in E$ and $r(x)=q(x)$ for all $x\in F$ Show that $r:E\sqcup F\to B$ is a covering space.

Take $x\in B$, then there is an open $U$ of $B$ such that $x\in U$ and $p^{-1}(U)=\sqcup_{\alpha\in A}V_{\alpha}$, where $V_{\alpha}\in E$ for all $\alpha\in A$, in addition, there is also an open $W$ of $B$ such that $x\in W$ and $q^{-1}(W)=\sqcup_{\beta\in B}S_{\beta}$, where $S_{\beta}\in F$ for all $\beta\in B$. But I do not know what else to do, could someone help me please? Thank you very much.

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So, $p^{-1}(U)=\sqcup_\alpha V_\alpha$ with $V_\alpha\subseteq E$ and such that $p|_{V_\alpha}:V_\alpha\to U$ is a homeomorphism.
Restricting the codomains to $U\cap W$, we still obtain homeomorphisms $V_\alpha\cap p^{-1}(W)\to U\cap W$,
and we will explicitly have $$r^{-1}(U\cap W)=\bigsqcup_\alpha (V_\alpha\cap p^{-1}(W))\ \sqcup\ \bigsqcup_\beta (S_\beta\cap q^{-1}(U))\,.$$