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Why is $E \sqcup F$ not path connected?

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 $q^{-1}(W)=\sqcup_{\beta\in B}S_{\beta}$

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
$$r^{-1}(U\cap W)=\bigsqcup_\alpha (V_\alpha\cap p^{-1}(W))\ \sqcup\ \bigsqcup_\beta (S_\beta\cap q^{-1}(U))\,.$$

That is from another question. But I am interested in how to show that $E \sqcup F$ is not path connected.


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