Suppose that the set $A \subset \mathbb{R}^2$ can be written as follows:
\begin{align}
A= \bigcup_{i \in I} S_i
\end{align}
where
\begin{align}
S_i =\{ x \in \mathbb{R}^2: \|x\|=c_i \}
\end{align}
for some $c_i>0$.
Now suppose that a set $A$ is nowhere dense in $\mathbb{R}^2$. Can we show that $I$ has at most countable cardinality?
I think you need to assume that, for $i,j\in I$ with $i\neq j$, it must hold that $c_i\neq c_j$. Otherwise, you can take all $c_i$'s to be equal and get an obvious counterexample. From now on, I suppose that the $c_i$'s are pairwise distinct.
Let $I\subseteq \mathbb{R}$ be the Cantor set. For each $i\in I$, we take $c_i$ to be $1+i$. Note that $I$ is uncountable and $A$ is a compact set (noting that $A$ can be identified as the product of two compact spaces: the Cantor set $I$ and the unit circle $\mathbb{S}^1$). Therefore, the closure $\bar{A}$ of $A$ in $\mathbb{R}^2$ is $A$ itself. Thus, the interior of $\bar{A}$ is the same as the interior of $A$, and $A$ has empty interior (why?). That is, $A$ is nowhere dense.