An example of a similar universal cover for 5 points.

I am not sure what happens with your definition, but here is an answer when we modify (2) to

(2') $A$ has Hausdorff dimension $\le 1$ (e.g. $A$ is a countable union of smooth curves).

Such sets $A$ have empty interior but the converse is false.

I will use the notation $S_n'$ for the collection of corresponding $n$-universal subsets of $R^2$ (by the way, you should not use the name "universal cover" for this, it is already taken).

Here is a proof that $S_n'=\emptyset$ for $n\ge 5$.

Consider the $n$-fold product $A^n$ of $A\subset R^2$ satisfying (2'); $A^n\subset R^{2n}$. The group $G$ of similarities of the Euclidean plane is a Lie group of dimension $4$. The group $G$ acts smoothly on $R^2$ and we obtain the corresponding diagonal action of $G$ on $R^{2n}$: $g(z_1,...,z_n)=(gz_1,...,gz_n)$, where $z_k\in R^2$ for each $k$. In other words, we have the smooth map $\mu: G\times R^{2n}\to R^{2n}$, $$ \mu(g, z_1,...,z_n)= (gz_1,...,gz_n). $$ The statement that $A$ satisfies your universality condition is equivalent to the property that $\mu(G\times A^n)=R^{2n}$. However, $\mu$ is locally Lipschitz (since it is smooth) $G$ has Hausdorff dimension $4$, $A^n$ has Hausdorff dimension $\le n$, hence, $\mu(G\times A^n)$ has Hausdorff dimension $\le 4+n$. For $n\ge 5$, $4+n< 2n$ and $2n$ is the Hausdorff dimension of $R^{2n}$. (For $n=4$ we have the equality.) Hence, $\mu(G\times A^n)$ cannot equal $R^{2n}$. Thus, $A$ cannot belong to $S'_n$.

The same argument works if (2') is replaced by

(2'') $A$ has Hausdorff dimension $<2$.

I let $S_n''$ denote the corresponding collection of universal subsets of $R^2$.

Then the result is that $$ \bigcap_{n\ge 1} S_n'' =\emptyset. $$

I do not know what happens for your original definition of $S_n$. One can attempt to replace Hausdorff dimension with topological dimension. Then $int(A)=\emptyset$ means that $dim(A)\le 1$, hence $dim(A^n)\le n$. However, there are examples of smooth maps which raise topological dimension. For instance, one can take a topological curve $P\subset {\mathbb R}^3$ (the graph of a Peano curve), such that for the action $\mu$ of ${\mathbb R}$ on ${\mathbb R}^3$ by translations along the $x$-axis, $\mu({\mathbb R}\times P)$ is 3-dimensional, equal to the product of the unit square and the real line.

Edit. The fact that $$ \bigcap_{n\in {\mathbb N}} S_n \ne \emptyset $$ is proven in Theorem 1.12 in

C. G. Wastun, Universal covers of finite sets. J. Geom. 32 (1988), no. 1-2, 192–201.

In fact, he proves even more: There exists a closed subset $A$ with empty interior in $E^2$ such that for each finite subset $F\subset E^2$ there exists a horizontal translation $T$ of $E^2$ such that $T(F)\subset A$. His sets $A$ are products of certain closed, perfect, totally disconnected subsets of the x-axis with the y-axis.