Let $p: \tilde{X} \rightarrow X$ be a covering map in the broad definition (empty sum allowed, as Hatcher does). First, $p[\tilde{X}]$ is open in $X$: let $y \in X$ with $x \in \tilde{X}$ such that $p(x) = y$. Let $U$ be an evenly covered neighbourhood of $y$, so $p^{-1}[U] = \sum_{i \in I} O_i$ (disjoint sum over index set $I$). As $x$ is in the sum, it is non-empty: some $O_{i_{0}}$ contains $x$ and $p |_{O_{i_0}}$ is a homeomorphism between $O_{i_{0}}$ and $U$. In particular $U \subset p[\tilde{X}]$, so $y$ is an interior point of $p[\tilde{X}]$.
Suppose $y \notin p[\tilde{X}]$. Then again take an evenly covered neighbourhood $U$ of $y$: $p^{-1}[U] = \sum_{i \in I} O_i$, disjoint sum over some index set $I$, where for every $i \in I$, the map $p|_{O_i}$ is a homeomorphism between $O_i$ and $U$. This means (!) that $I = \emptyset$, as otherwise we'd have a preimage for $y$, contradicting how we picked $y$. So in fact $p^{-1}[U] = \emptyset$, and this shows that $U \subset X\setminus p[\tilde{X}]$, so $p[\tilde{X}]$ is closed.
Now if $X$ is connected and $\tilde{X}$ is non-empty, then $p[\tilde{X}]$ is closed, open and non-empty, so equals $X$ by connectedness. So $p$ is surjective.
These are the only conditions we need: $X$ connected and $\tilde{X} \neq \emptyset$.
Let $X$ and $Y$ be the two graphs having two vertices and three edges. There is a common two-sheeted covering space $Z$ of $X$ and $Y$ which is a graph with four vertices and six edges. Exercise: Find $Z$ and show that $X$ and $Y$ are not covering spaces of any other spaces (besides themselves, of course).
Interesting side note: The covering spaces $Z\to X$ and $Z\to Y$ are defined by free actions of ${\mathbb Z}_2$ on $Z$. These generate a ${\mathbb Z}_2\times{\mathbb Z}_2$ action, but it is not a free action since the product of the two generators has a fixed point.
Best Answer
When a topology is specified by giving a basis, an open set in the basis is called a basic open set.
That's not what he is saying. $X$ is a graph, and $\tilde{X}$ is some topological covering space. He wants to prove that the graph structure on $X$ gives a graph structure on $\tilde{X}$. He then does so, but then wants to show that this graph structure's topology on $\tilde{X}$ is the same as the original topology on $\tilde{X}$ (a set can have several different topologies: he is saying these two are the same here). Hence the graph structure respects the original topology on $\tilde{X}$. This happens because the graph structure is induced by $X$ and the local covering conditions lift the right way upstairs. He is not claiming that this means that $X$ and $\tilde{X}$ have the same topology.