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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.

Thanks