I like the following abstract construction, which relies on a connection between quasi-continuity and the fine topology and also works in the setting of $p\ne 2$. It is debatable whether the resulting example is “simple”. The standard reference for the required techniques is Adams, Hedberg: Function spaces and potential theory, 1995.

Let $(B_k)$ be a sequence of open balls contained in the open unit ball $D\subset\mathbb{R}^n$ such that $A:=\bigcup_{k} B_k$ is dense in $\overline{D}$, but such that $\operatorname{cap}(A)<\operatorname{cap}(D)$. This can be arranged by centering the balls in a countable dense subset of $D$ and letting the radii decrease sufficiently rapidly. [As we are in the Hilbert space setting $p=2$, it is here where we need $n\ge2$.] Let $K:=\overline{D}\setminus A$. Note that $K$ is compact and has no interior points, but is necessarily still large in measure.

By Theorem 11.3.2 (in Adams, Hedberg), there exists a nontrivial nonnegative Sobolev function $u$ supported in $K$. In fact, let $w\in H^1(\mathbb{R}^n)$ be the capacitary extremal of $A$. Then $0\le \tilde{w}\le 1$ and $\tilde{w}=1$ quasi-everywhere on $A$, where $\tilde{w}$ is the quasi-continuous representative of $w$. As $\operatorname{cap}(A)<\operatorname{cap}(D)$, we cannot have $w=1$ a.e. on $D$. Let $\varphi\in C^\infty_c(\mathbb{R}^n)$ be such that $\varphi(x)>0$ for all $x\in D$ and $\varphi(x)=0$ for all $x\notin D$. Set $u := \varphi(1-w)$. Then $u\in H^1(\mathbb{R}^n)$ has the desired properties.

It follows that $U := \{x\in\mathbb{R}^n: \tilde{u}(x)>0\}$ is quasi-open and (possibly after removing a polar set) $U\subset K$. Let $O$ be the fine interior of $U$. Then $O\subset U$ and $U\setminus O$ is polar; see Section 6.4 and Proposition 6.4.12 (in Adams, Hedberg). As $u$ is nontrivial, both $U$ and $O$ have positive measure.

Consequently $O$ is a nontrivial finely open set that has no Euclidean interior (as it is contained in $K$). Let $V$ be an open set. If $V\cap O$ is nonempty, then $V\cap A$ has positive measure and clearly $V\cap A\subset V\setminus O$. Otherwise [i.e. if $V\cap O=\emptyset$] the set $O\setminus V=O$ has positive measure. It follows that the symmetric difference of $O$ and $V$ cannot be polar.
So $O$ is not equivalent (up to a polar symmetric difference) to any open set.