$∂(A\cup B)$ is a subset of $∂A \cup ∂B$ Please let me know if you think my proof of the above is correct. ($∂A$ denotes the boundary of $A$).
Suppose $\vec{x}\in ∂(A\cup B)$. Then $\vec{x}\in \overline{A \cup B}=\bar{A}\cup\bar{B}$ and $B_\varepsilon(\vec{x}) \cap (\bar{A}\cap \bar{B})\ne \emptyset\ne B_\varepsilon(\vec{x})\cap (\bar{A}\cup\bar{B})^c$. All the above implies that $\vec{x}\in ∂A$ or $\vec{x}\in ∂B$. Thus $\vec{x}\in ∂A\cup ∂B$.
I'm in a little bit of a doubt about the rigor of this proof.

Yes, it is right. A better proof is as 
\begin{align}
\partial(A\cup B)&=\overline{A\cup B}\cap \overline{(A\cup B)^c}
\\
&=(\overline{A}\cup \overline{B}) \cap \overline{A^c\cap B^c}
\\
&=(\overline{A}\cap \overline{A^c\cap B^c})\cup (\overline{B} \cap \overline{A^c\cap B^c})
\\
&\subset(\overline{A}\cap \overline{A^c})\cup (\overline{B} \cap \overline{B^c})
\\
&=\partial A\cup \partial B
\end{align}
