Let $A, B \subseteq ℂ$, if $A \subseteq B$, then $\partial A \subseteq \partial B$? Let $A, B \subseteq ℂ$, if $A \subseteq B$, then $\partial A \subseteq \partial B$?
I thought of a counter example but I'm not sure if the boundaries ($\partial$) are context dependent.
Here it is:
$A = ℝ, B = ℂ$
So obviously, $A \subseteq B$, but here's the problem I'm having.
I know that ℝ is both open and closed set (?), and it's boundaries should be an empty set - but does the context of the topology, that since we look at ℝ as $ℝ \subseteq ℂ$, it's boundaries is just ℝ? Thus satisfies a counter example for the said claim?
 A: $\mathbb{R}$ is only clopen in its own topology, not necessarily any other. In general, for a topology defined on a set $S$, $S$ as a set is a clopen subset, but that need not extrapolate to any supersets of $S$ in their own respective topologies.
It's not hard to see that $\mathbb{R}$ is not open in the grander context of $\mathbb{C}$. Take any ball with center in $\mathbb{R}$ and positive radius, and there will be a non-real complex number contained inside (say, just above the center).

Anyhow, as a simpler counterexample, just try two circles with differing radii contained in each other, like below:

A: $A, B\subset \Bbb{C}$ and $A\subset B$
Then $\overline{A}\subset \overline{B}$.
If $\partial{A}\subset\partial{B}$ then $\overline{A}\setminus \overset{o}A\subset\overline{B}\setminus \overset{o}B$
implies $\overset{o}B\subset \overset{o}A$
$A\subset B$ doesn't imply $\overset{o}B\subset \overset{o}A$.
Counter examples : Choose $A\subset B $ such that $\overset{o}B\not\subset \overset{o}A$.
$A=C_0(1) :\{z:|z|=1\}$ , $A=C_0(2) :\{z:|z|=2\}$
Then $A\subset B$ but $\overset{o}B=D(0, 2) $ and $\overset{o}A=D(0, 1) $.

Now you can see why your example won't work.
$A=\Bbb{R}\subset \Bbb{C}$ is nowhere dense. It's interior is empty! Empty set is subset of every set. So any set containing $\Bbb{R}$ won't provide a counter example.
