Prove that the projection map $\rho:\mathbb R× \{0, 1\} → \mathbb R: \space \rho(a, i)=a$ is a covering map. This question seems easy if we consider the $\{0,1\}$ is some topology where singletons are open (i.e. discrete topology). Since in this case, for $(x,y) \in \mathbb R× \{0, 1\}$, we can pick the open set $U = (x- \epsilon, x+\epsilon)×\{y\}$ for $\epsilon > 0$ and check that $\rho$ is a local homeomorphism. 
However, there was no extra information in the question text. Am I missing something here, or is the question only true if we can choose the topology in $\{0,1\}$ (for instance, trivial topology would not work)?
 A: Well, for $x\in\mathbb R$, you can actually take the whole $\mathbb R$ as a neighborhood, then the reciprocal image is $\mathbb R\times\{0,1\}$ which is indeed homeomorphic to $\mathbb R\sqcup\mathbb R$. There's nothing wrong about it, this is called a ($2$-sheeted) trivial covering space.
A: More generally, a projection map $p : X \times F \to X, p(x,f) = f$, where $X$ is an arbitrary space, is a covering projection if and only if $F$ is discrete.
If $F$ is discrete, then $p^{-1}(X) =  \bigcup_{f \in F} X \times \{ f \}$ is a partition into pairwise disjoint open subsets $X \times \{f \} \subset X \times F$ which are mapped by $p$ homeomorphically onto $X$.
If $p$ is a covering projection, then each $x \in X$ has an open neighborhood $U \subset X$ such that $p^{-1}(U) = \bigcup_{\alpha \in A} V_\alpha$ with a family $(V_\alpha)_{\alpha \in A}$ of pairwise disjoint open subsets of $X \times F$ which are mapped by $p$ homeomorphically onto $U$. Let $f \in F$. Then $(x,f) \in p^{-1}(U)$ and there exists a unique $\alpha \in A$ such that $(x,f) \in V_\alpha$. We conclude that $W_\alpha = p^{-1}(x) \cap V_\alpha$ is open in $p^{-1}(x) =  \{ x \} \times F$ and is mapped by $p$ homeomorphically onto $p(W_\alpha) = \{ x \}$. Hence $W_\alpha$ contains exactly one point, thus $W_\alpha = \{ (x,f) \}$. Therefore $\{ (x,f) \}$ is open in $ \{ x \} \times F$ which implies that $\{ f \}$ is open in $F$.
In your text it must therefore be assumed that $\{ 0, 1 \}$ has the discrete topology. However, I believe it the standard assumption that any topological space with finitely many points finite carries the discrete topology if nothing else is explicitly stated.
