If $A\subset B$ then $\bar{A}\subset \bar{B}$ Please let me know if you find my proof satisfactory.
Let $A\subset B\subset \mathbb{R}^n$, where $\bar{A}$ is the closure of $A$.
If $\vec{x}\in \bar{A}$ then $\vec{x}\in A^\circ$ or $\vec{x}\in ∂A$ (in the boundary of $A$). If $\vec{x}\in A^\circ$ then $\vec{x}\in A \implies \vec{x}\in B\subset \bar{B}$.
If $\vec{x} \in ∂A$ then $B_\varepsilon(\vec{x})\cap A\ne \emptyset \ne B_\varepsilon(\vec{x})\cap B$ (for any $\varepsilon > 0$). Thus, since $A\subset B$, $\vec{x}\in B \implies \bar{A}\subset \bar{B}$.
 A: $A\subset B\implies A\subset\bar B\implies\bar A\subset\bar B $.
A: For $x\in \partial A$ you cannot proof in general that $x \in B$. Also $B_\varepsilon(x)\cap A\neq\emptyset\neq B_\varepsilon(x) \cap B$ is not realy an argumentation. The logic should be $B_\varepsilon(x)\cap A\neq\emptyset \Rightarrow B_\varepsilon(x)\cap B\neq\emptyset$ since  $B_\varepsilon(x)\cap A\subseteq  B_\varepsilon(x)\cap B$ for all $\varepsilon>0$. As I said, this cannot imply that $x \in B$, but it does imply that either $x \in \partial B$ or $x \in \mathring{B}$ which is sufficient for you.
That comment was right of course, I fixed the answer.
A: Almost.
If $x\in∂A$ then, for any $\epsilon>0$ we have $$B_{\epsilon}(x)\cap A\neq\emptyset\Rightarrow B_{\epsilon}(x)\cap B\neq\emptyset$$ and $$B_{e}\cap A^{c}\neq\emptyset$$ where $A^{c}$ is the complement of $A$. If $B_{\epsilon}(x)\cap B^{c}\neq\emptyset$ we have $x\in ∂B$. If $B_{\epsilon}(x)\cap B^{c}=\emptyset$ we have $x\in B^\circ$. Therefore $x\in\overline{B}$.
