The answer to the second question is yes: there is an arc containing uncountable points of $B\cap\partial C$. It is enough to prove it in the case $n=2$.

Applying an affine transformation if necessary, we can suppose that $[0,1]^2\subseteq B$, $[0,1]\times\{0\}\subseteq C$ and $[0,1]\times\{1\}\subseteq D$. This implies that for any $x\in[0,1]$, there is some point of $\partial C$ in $\{x\}\times[0,1]$.

Now let $W$ be the set of finite strings of $0$ and $1$. Given a word $w\in W$, we write $w0$ and $w1$ for the words obtained by adding "$0$" or "$1$" at the end of $w$.

To each word $w\in W$ we will associate a rectangle $R_w=[x_w,x_w']\times[y_w,y_w']\subseteq[0,1]^2$ such that:

- For any $w\in W$, the set $R_w\cap\partial C$ has an uncountable projection onto the $x$-axis.
- If $w$ has length $n$, then $R_w$ has diameter $\leq2^{-n}$.
- For any $w\in W$, $R_{w0}$ and $R_{w1}$ are contained in $R_w$, and $x_{w0}'<x_{w1}$. So $R_{w0}$ and $R_{w1}$ are disjoint.

It is easy to see how to construct the rectangles inductively. Now let $\omega\in2^\mathbb{N}$ be an infinite word of ones and zeros, with $\omega_n$ being the finite word formed by the first $n$ characters of $\omega$. Remember that $2^\mathbb{N}$ (with the product topology) is homeomorphic to the ternary Cantor set $X\subseteq[0,1]$ via the function $f:2^\mathbb{N}\to X;\omega=(x_n)_{n\in\mathbb{N}}\mapsto\sum_{n\in\mathbb{N}}2x_n3^{-n}$.

To each $\omega\in2^\mathbb{N}$ we associate the point $p_\omega=\cap_{n\in\mathbb{N}}R_{\omega_n}$. This defines an imbedding $f:X\to\partial C$, because $f$ is continuous from a compact space to a T2 space and $f$ is bijective: in fact, if $x_1,x_2\in X$ with $x_1<x_2$, then the $x(f(x_1))<x(f(x_2))$ (where $x(p)$ represents the $x$-coordinate of a point $p$).

We can extend this homeomorphism to an arc $F:[0,1]\to[0,1]^2$: to do this, we just have to define $F$ in the countable intervals $(p_n,q_n)$ of $[0,1]\setminus X$. We do this by interpolating linearly between $f(p_n)$ and $f(q_n)$, that is, $F(tp_n+(1-t)q_n)=tf(p_n)+(1-t)f(q_n)$. The continuity of $F$ can be deduced easily from the continuity of $f$, and $F$ is injective because different points of $[0,1]$ get sent to points of $[0,1]^2$ with different $x$-coordinates.

So this arc contains continuum many points of $B\cap\partial C$.