A non-open evenly covered subset?

Question

I have the following definition of an evenly covered subset:

Let $p\colon\tilde X\to X$ be a subjective continuous map. Then a subset $U$ (not necessarily open) is said to be evenly covered by $p$ iff $p^{-1}(U)$ can be written as a disjoint union $\bigsqcup_\alpha \tilde U_\alpha$ of open sets $\tilde U_\alpha$'s in $\tilde X$, each homeomorphic to $U$ via $p$'s restriction $\tilde U_\alpha\to U$.

I was convinced that any evenly covered set $U$ would trivially be open, for it's "just like" (via restricted homeomorphisms) any of the copies $\tilde U_\alpha$'s, which are all open in $\tilde X$. However, when I tried to pen down a proof of this plausible statement, I got stuck, and thus am now doubting the truth of it.

Can you think of a counterexample? (Or of a proof?)

Answer 1

Let $X$ be some space, and $Y$ be $X$ with discrete topology. Then $p:Y\to X$ defined by $p(x) = x$ for $x\in Y$ is continuous and surjective.

If $A\subseteq X$ is discrete, then $A$ is evenly covered by $p$.

For example, take $X = \mathbb{R}$ and $A = \mathbb{N}$.