Let $p:E\rightarrow X$ be a covering map. Let $V$ be any open subset of $E$ and $e$ be any element of $V$. I feel that the following statement must be true:
There exists an evenly covered open subset $U$ of $X$ such that $p^{-1}[U]=\cup_{i\in I} S_i$ ($\{S_i\}_{i\in I}$ are the sheets over $U$) and $e\in S_{i_0}\subseteq V$ for some $i_0\in I$.
I spent some time trying to prove it, but I didn't succeed . I prefer hints more than full answers (if the statement is true).
Thank you
On
Let $U\subset X$ be an evenly covered neighbourhood of $p(e)$ and let $U_0=U\cap p(V)$. We know that $e\in V$ so $p(e)\in p(V)$ and so $p(e)\in U_0$. Also $U_0\subset U$ and so $U_0$ is also an evenly covered neighbourhood of $p(e)$.
Because $U_0\subset p(V)$ we have that $p^{-1}( U_0)\cap V\neq\emptyset$ and indeed this means that at least one of the sheets of $U_0$ is contained in $V$ and exactly one of these contains $e$. This sheet is equal to $S_i$ for some $i\in I$ by the definition of $p^{-1}(U_0)$ for $U_0$ evenly covered.
(Edited. The objection of the OP was substantial.)
Let $e':=p(e)\in X$. The point $e'$ has an open neighborhood $U'$ which is evenly covered by $p^{-1}(U')\subset E$. Let $S_0$ be the sheet of $p^{-1}(U')$ that contains the point $e$. Then $W:=S_0\cap V$ is an open neighborhood of $e$, and $W':=p(W)\subset U'$ is an open neighborhood of $e'$. This neighborhood is evenly covered by $p^{-1}(W')$, and the sheet of $p^{-1}(W')$ containing $e$ is $W\subset V$.