For subsets $A, B \subseteq \mathbb{R}^{n}$, is it true that $A \subseteq B \implies \partial A \subseteq \partial B$? 
For subsets $A, B \subseteq \mathbb{R}^{n}$, is it true that $A
 \subseteq B \implies \partial A \subseteq \partial B$?

I believe the answer is no. But, I cannot come up with a counterexample. I need the boundary of $A$ to not be contained in the boundary of $B$, but I need $A$ to be in $B$.
I was thinking like two concentric circles with different radii won't have intersecting boundaries. I don't know how to write them as sets in set-builder notation though. 
 A: Take $B=\mathbb R^{n}$. then $\partial B=\emptyset$. Can you think of some subset whose boundary is not empty?. 
A: 
I believe the answer is no.

That is correct. The boundaries can even be disjoint.

I was thinking like two concentric circles with different radii won't have intersecting boundaries.

Correct again, assuming that with “circles” you mean $n$-dimensional spheres.

I don't know how to write them as sets in set-builder notation though.

I would write it like this: A counter example are the concentric spheres
$$
 A = \{ x \in \Bbb R^n: \Vert x \Vert < 1 \} \\
 B = \{ x \in \Bbb R^n: \Vert x \Vert < 2 \}
$$
with the disjoint boundaries
$$
 \partial A = \{ x \in \Bbb R^n: \Vert x \Vert = 1 \} \\
 \partial B = \{ x \in \Bbb R^n: \Vert x \Vert = 2 \}
$$
More generally, if the closure of $A$ is contained in the interior of $B$ then $\partial A \cap \partial B = \emptyset$.
A: Another idea: let $B$ a ball and $A$ a dense subset of $B$. Then $\partial B$ is simply the sphere but $\partial A$ is much bigger.
