Suppose that $B \subseteq A$. Then by using that $\overline{A_1\cup A_2} = \overline{A_1}\cup\overline{A_2}$ for any sets $A_1$ and $A_2$,
$$
\partial A\cup\partial B
\subseteq \overline{X\setminus A}\cup\overline{B}
= \overline{(X\setminus A)\cup B}
= \overline{X\setminus(A\setminus B)}. \tag{1} $$
Now we also assume that $\overline{B}\subseteq\mathring{A}$. Then
$$ \overline{X\setminus A} \cap \overline{B}
\subseteq \overline{X\setminus A} \cap \mathring{A}
= \varnothing. $$
So
\begin{align*}
\partial B
&= \overline{X\setminus B} \cap \overline{B} \\
&= (\overline{A\setminus B} \cup \overline{X\setminus A}) \cap \overline{B} \\
&= (\overline{A\setminus B} \cap \overline{B}) \cup (\overline{X\setminus A} \cap \overline{B}) \\
&= \overline{A\setminus B} \cap \overline{B}
\end{align*}
and this shows that $\partial B \subseteq \overline{A\setminus B}$. Simiarly,
\begin{align*}
\partial A
&= \overline{X\setminus A} \cap \overline{A} \\
&= \overline{X\setminus A} \cap (\overline{A \setminus B} \cup \overline{B}) \\
&= (\overline{X\setminus A} \cap \overline{A \setminus B}) \cup (\overline{X\setminus A} \cap \overline{B}) \\
&= \overline{X\setminus A} \cap \overline{A \setminus B}
\end{align*}
shows that $\partial A \subseteq \overline{A\setminus B}$. Consequently
$$ \partial A \cup \partial B \subseteq \overline{A\setminus B} \tag{2} $$
and combining $\text{(1)}$ and $\text{(2)}$ proves the inclusion $\partial A \cup \partial B \subseteq \partial (A\setminus B)$.
Yes, both hold.
For any family $\{H_i\}_{i\in I}$ of subsets (not necessarily subgroups) of $G$, $\langle H_i\rangle$ is the subgroup $K$ that satisfies the following two conditions:
- $H_i\subseteq K$ for each $i\in I$; and
- if $M$ is any subgroup of $G$ such that $H_i\subseteq M$ for every $i\in I$, then $K\subseteq M$.
This is the "top-down description" of the subgroup generated. You use these properties in the second displayed equation after $(2)$.
We want to show that
$$\bigl\langle G_1\cup G_2\cup G_3\bigr\rangle = \Bigl\langle \bigl\langle G_1\cup G_2\bigr\rangle\cup G_3\Bigr\rangle.$$
(We don't even need to assume these are subgroups).
$\subseteq)$ Note that
$$\begin{align*}
G_1\cup G_2\subseteq \langle G_1\cup G_2\rangle &\subseteq \Bigl\langle \bigl\langle G_1\cup G_2\bigr\rangle\cup G_3\Bigr\rangle\\
G_3\subseteq \langle G_1\cup G_2\rangle\cup G_3 &\subseteq \Bigl\langle \bigl\langle G_1\cup G_2\bigr\rangle\cup G_3\Bigr\rangle
\end{align*}$$
Therefore,
$$(G_1\cup G_2)\cup G_3 = G_1\cup G_2\cup G_3 \subseteq
\Bigl\langle \bigl\langle G_1\cup G_2\bigr\rangle \cup G_3\Bigr\rangle,$$
and since $\langle G_1\cup G_2\cup G_3\rangle$ is the smallest subgroup that contains $G_1\cup G_2\cup G_3$, we have
$$\langle G_1\cup G_2\cup G_3\rangle \subseteq \Bigl\langle\bigl\langle G_1\cup G_2\bigr\rangle\cup G_3\Bigr\rangle.$$
$\supseteq)$ Since $G_1\cup G_2\subseteq G_1\cup G_2\cup G_3\subseteq \langle G_1\cup G_2\cup G_3\rangle$, we have
$\langle G_1\cup G_2\rangle \subseteq \langle G_1\cup G_2\cup G_3\rangle$.
We also have $G_3\subseteq G_1\cup G_2\cup G_3\subseteq \langle G_1\cup G_2\cup G_3\rangle$.
Since both $\langle G_1\cup G_2\rangle$ and $G_3$ are contained in $\langle G_1\cup G_2\cup G_3\rangle$, their union is contained there as well, so
$$\langle G_1\cup G_2\rangle\cup G_3\subseteq \bigl\langle G_1\cup G_2\cup G_3\bigr\rangle,$$
and therefore
$$\Bigl\langle \bigl\langle G_1\cup G_2\bigr\rangle \cup G_3\Bigr\rangle \subseteq \bigl\langle G_1\cup G_2\cup G_3\bigr\rangle,$$
as desired. $\Box$
More generally, for any family $\{B_i\}_{i\in I}$ of subsets of $G$, and any partition $\{I_j\}_{j\in J}$ of $I$ (so $\cup_{j\in J}I_j=I$), we have
$$\left\langle \bigcup_{i\in I}B_i\right\rangle = \left\langle \bigcup_{j\in J}\left\langle \cup_{i\in I_j}B_k\right\rangle\right\rangle.$$
As to whether your description of $(1)$ is accurate, note that in general if $S$ is a subset of $G$, $\langle S\rangle$ is the collection of all finite products of elements of $S$ and their inverses. But if $S$ is closed under inverses, that is, if $S$ satisfies the condition
$$\text{if }s\in S,\text{ then }s^{-1}\in S,$$
then this is exactly the same thing as the collection of all finite products of elements of $S$. (Note the empty product is a finite product and gives the identity element). Since your $\cup\mathcal{H}$ is a union of subgroups, it is certainly closed under inverses; and you can ignore the empty product since the set already contains $e$; so your description of $\langle \cup\mathcal{H}\rangle$ is accurate if the family of subgroups is nonempty.
Best Answer
This is true if and only if $V \subseteq W$. If this is true, then the second subset relationship implies the result. If this is not true, then $U$ must contain all elements of $V,$ including those outside $W,$ so the statement is false.