Preimages of evenly covered basis gives a basis? Let $(X, \mathcal{T}_X), (Y, \mathcal{T}_Y)$ be topological spaces, and let $p : Y → X$ be a covering map. Then there exists some open cover $\mathcal{U} \subseteq \mathcal{T}_X$ such that every $U ∈ \mathcal{U}$ is evenly covered by $p$, i.e. for every $U∈ \mathcal{U}$ there is a set $\mathcal{V}_U \subseteq \mathcal{T}_Y$ of pairwise disjoint open sets such that $p^{-1}U = \sqcup_{V ∈ \mathcal{V}_U}V$, and that $p|_V : V \tilde{→} U$ is a homeomorphism for each $V ∈ \mathcal{V}_U$.
Now, I have seen the following claim going around which in my understanding should be wrong (although it probably isn't).
The first part of the claim is there exists an evenly covered basis $\mathcal{B}_X$ for $X$; the second part is that $\bigcup\limits_{U ∈ \mathcal{B}_X} \mathcal{V}_U$ forms a basis for $\mathcal{T}_Y$.
Consider the topological space $(X, \{\emptyset, X_1, X_2, X\})$. Suppose $X_1 ∪ X_2 = X$, but they are not disjoint. Then $\mathcal{U} = \{X_1, X_2\}$ is an open covering of $X$. Say that we have $\mathcal{V}_{X_1} = \{V_1, V_2, V_3 \}$ and $\mathcal{V}_{X_2} = \{W_1, W_2, W_3 , W_4\}$ so that $\mathcal{U}$ is in fact an evenly covered basis for $\mathcal{T}_X$. In the definition of covering spaces, there is no demand on the $V_i$ to not-overlap with a $W_j$. Note that as $X_1 ∩ X_2 ≠ \emptyset $, we have  $\emptyset \neq p^{-1}(X_1 ∩ X_2) = p^{-1}X_1 \cap  p^{-1}X_2 = \sqcup_{i=1}^3 V_i \cap \sqcup_{i=1}^4 W_i$.  So at least some $V_i$ overlaps with some $W_j$. Say, $V_1 ∩ W_1 ≠ \emptyset$. $V_1$ and $W_1$, however, being open sets in $Y$, must result in $V_1 ∩ W_1 ∈ \mathcal{T}_Y$. But this is in contradiction with the claim that $\bigcup\limits_{U ∈ \mathcal{B}_X} \mathcal{V}_U$ forms a basis for $\mathcal{T}_Y$, because we can never get the open set $V_1 ∩ W_1$ by taking unions of elements not strictly smaller than $V_1$ and $W_1$!
What's going on here?
 A: It is wrong. We shall use notation (e.g. sheet structure and sheet number) and basic results from Covering projections: What are the sheets over an evenly covered set?
The claim says that

*

*There exists an evenly covered basis $\mathcal B_X$ for $X$.


*For each $U \in \mathcal B_X$ let $S(U)$ be a sheet structure over $U$. Then $\mathcal{B}_X^Y = \bigcup\limits_{U \in \mathcal{B}_X} S(U)$ forms a basis for $Y$.
Note that in general the sheet structure over $U$, i.e. the decomposition of $p^{-1}(U)$ into sheets, is not unique. Non-uniqueness always occurs if $U$ is not connected with sheet number $> 1$. In that case we actually have to choose a sheet structure over $U$. This means that $\mathcal{B}_X^Y$ in general is not uniquely determined by $\mathcal{B}_X$, but involves a choice for each non-connected $U \in \mathcal{B}_X$ with sheet number $> 1$.
Now let $X = \{0\} \cup \{1/n \mid n \in \mathbb N\}$ with the subspace topology inherited from $\mathbb R$ and $Y = X \times \mathbb Z$. Let $p$ be the projection; this is a covering map. Since $X$ is evenly covered, also each open subset of $X$ is evenly covered.
Let $\mathcal B_X$ be any basis for $X$. Each $U \in \mathcal B_X$ has a maximal element $x_U \in U$ with respect to the natural order of $X$. Of course $x_U > 0$ since $\{0\}$ is not open.
For $k \in \mathbb Z$ define $V_k(U) =( U \setminus \{x_U\}) \times \{k\} \cup \{(x_U,k+1)\}$. Then the $V_k(U)$ form a decomposition of $p^{-1}(U)$ into sheets.
We claim that the collection $\mathcal B_X^Y  =\{V_k(U) \mid U \in \mathcal B_X, k \in \mathbb Z\}$ is not a basis for $Y$.
The set $X \times \{0\}$ is an open neigborhood of $(0,0)$ in $Y$. If $\mathcal B_X^Y$ were a basis for $Y$, then $X \times \{0\}$ should contain some $V_k(U)$ with $(0,0)  \in V_k(U)$. The latter is satisfied iff $k = 0$ and $0  \in U$. However, $(x_U,0)$ is not contained in $V_0(U)$. This is a contradiction.
So what can be said positively?

If $X$ is locally connected, then it has a basis $\mathcal B_X$ of evenly covered open connected sets. For each such basis, $\mathcal B_X^Y$ is a basis for $Y$. In particular, $Y$ is locally connected.

Note that each evenly covered open connected set $U$ has a unique sheet structure $S(U)$ over $U$, thus no choice of a decomposition of $p^{-1}(U)$ into sheets is involved. Therefore in the above case $\mathcal B_X^Y$ is uniquely determined by $\mathcal B_X$.
Let $W$ be an open neigborhood of $y \in Y$. Let $U' \subset X$  be an evenly covered open set such that $p(y) \in U'$. Choose any sheet structure $S(U')$ over $U'$. There exists a unique $V'_y \in S(U')$ such that $y \in V'_y$. The set $W' = W \cap V'_y$  is an open neigborhood of $y$ and $U'' = p(W') \subset U'$ is an open neigborhood of $p(y)$. We have $W' \in S(U') \mid_{U''}$. There exists $U \in \mathcal B_X$ such that $p(y) \in U \subset U''$. The restriction $(S(U') \mid_{U''}) \mid_U = S(U') \mid_U$ is the unique sheet structure $S(U)$ over the connected $U$. Let $V = W' \mid_U \in S(U)$. Then $y \in V = W' \mid_U \subset W' =  W \cap V_y \subset W$ and  we are done.
Another positive result is this. For an open evenly covered $U \subset X$ let $S^*(U)$ be union of all sheet structures over $X$ (in other words, it is set of all open $V \subset p^{-1}(U)$ which are plain over $U$).

Let $\mathcal B_X$ be a basis of $X$ consisting of evenly covered open sets. Then $\bigcup\limits_{U \in \mathcal{B}_X} S^*(U)$ forms a basis for $Y$.

The proof is almost the same as the above proof for locally connected $X$. We get again $y \in V = W' \mid_U \subset W' =  W \cap V_y \subset W$, but now $V \in (S(U') \mid_{U''}) \mid_U \in S^*(U)$.
