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 $V \in (S(U') \mid_{U''}) \mid_U \in S^*(U)$.
Points 1. - 3. are trivial.
Proof of 4. :
If $U$ is not connected, then we have a decomposition $U = U_1 \cup U_2$ with nonempty disjoint open $U_i$. Let $S(U)$ be a sheet structure over $U$. Then we obtain sheet structures $S(U_i) = S(U) \mid_{U_i}$ over $U_i$. Now 1. applies to give distinct sheet structures over $U$. Conversely, let $U$ be connected. Let $S(U)$ be a sheet structure over $U$. Then each $V \in S(U)$ is connected, and moreover $V$ is a maximal connected subset of $p^{-1}(U)$ because any bigger connected $C \supset V$ would meet the open set $V^* = \bigcup_{V' \in S(U), V' \ne V} V'$. This would split $C$ into the disjoint nonempty open subsets $V$ and $C \cap V^*$. Thus each $V \in S(U)$ is a component of $p^{-1}(U)$. Since $\bigcup_{V \in S(U)} V = p^{-1}(U)$, we conclude that each component of $p^{-1}(U)$ is contained in $S(U)$.
Proof of 5. :
Let $S(U)$ be a sheet structure over $U$. Then all $W_V = W \cap V$, $V \in S(U)$, are open subsets of $W$ (some may be empty) which cover $W$ and are mapped by the homeomorphism $p_W : W \to U$ onto open $U_V \subset U$. Note that $p_W(W \cap V) = p_V(W \cap V)$. The $U_V$ cover $U$ and are pairwise disjoint because the $W_V$ are pairwise disjoint. Let $A = \{ V \in S(U) \mid W_V \ne \emptyset \}$. All $U_V$ with $V \in A$ are evenly covered with sheets structures $S(U) \mid_{U_V}$. For each $V \in A$ the restriction map $\rho_V : S(U) \to S(U) \mid_{U_V}$ is a bijection such that $\rho_V(V) = p_V^{-1}(U_V) = p_W^{-1}(U_V) = W_V$. Now fix $V_0 \in A$ and let $\phi_V : S(U) \to S(U) \mid_{U_V}$ be the bijection agreeing with $\rho_V$ for $V' \ne V,V_0$ and satisfying $\phi_V(V_0) = V, \phi_V(V) = V_0$. Then $W = \cup_{V \in A} \phi_V(V_0) \in S(U,\phi_V)$.
Best Answer
Because $p:C\to X$ is surjective and continuous, every open set $U$ in $X$ is covered by its preimage $p^{-1}(U)$ (which is open): here covered means that $p(V)=U$. The adjective evenly is used to mean that $p^{-1}(U)$ is a disjoint union of open sets $U_i$, all homeomorphic to $U$ under $p$; that is, each $p:U_i \longrightarrow U$ an homeomorphism.