Is this set countable? My motivation for this question comes from this earlier question and its answer. 
Let $\{E_i\}_{i=1}^{\infty}\subset\mathbb{R}^2$ be such that for each $i\in\mathbb{N}$,  $E_i$ is closed with non-empty interior, that is, $\text{Int}(E_i)\neq\emptyset$. Set
\begin{equation}
E=\bigcup_{i\in\mathbb{N}}E_i.
\end{equation}If $E$ is not dense in $\mathbb{R}^2$ and $\mathcal{L}^2(\partial E)\neq 0$, then is $\partial E\setminus\cup_{i\in\mathbb{N}}\partial E_i$ countable? Here, $\mathcal{L}^2$ denotes the two-dimensional Lebesgue measure. 
 A: False. Here is a counterexample. Let $E_i$ be a sequence of closed disks with center at the origin, where $E_i$ has radius $2-1/i$. 
Then the $E_i$ is an increasing sequence of closed disks approaching the disk with radius 2 at the origin, and hence $E$ is the open disk with radius 2 at the origin.
So $\partial E$ is the circle with radius 2 at the origin, and is trivially disjoint from $\partial E_i$, which is the circle of radius $2-1/i$ centered at the origin. So $\partial E\backslash \bigcup \partial E_i=\partial E$ is uncountable.
A: Let $E_i=[\frac1i,2]\times[0,1].$ Then
$$\partial\bigcup_{i\in I}E_i\setminus\bigcup_{i\in I}\partial E_i=\{0\}\times[0,1].$$
A: Let $\ E_n\ :=\ [0;\frac {n-1}n]^2\ $ for $\ n=1\ 2\ \ldots.\ $ Then
$\ \delta E\setminus\bigcup_{n=1}^\infty \delta E_n\ $ is not countable.
A: The previous answers by @bof and me are similar. I'll add another similar answer but I feel that it is more cute (a bit more elegant):
Let's consider disks:
$$ E_n\ :=\ \{ p\in\mathbb R^2: |p| \le 1-\frac 1{n+1} \} $$
Then $\ \delta E\setminus\bigcup_{n=1}^\infty\delta E_n\ $ is not countable.
