Boundary of union of open subsets Let $X$ be a topological space and each $V_i \subset X$ be an open subset of $X$, where $i \in I$. Denote $V_I = \{V_i : i \in I\}$. Below I'll show that
$$(*) \quad \quad \partial \left(\bigcup V_I\right) = \overline{\bigcup \partial V_I} \setminus \bigcup V_I $$
provided


*

*$X$ is locally connected, or

*$V_I$ is locally finite.


The problem
Does this equation hold in arbitrary topological spaces without restrictions? Also of interest are other conditions under which this equation holds.
Note
For an arbitrary collection $V_I$ in an arbitrary topological space,
$$\overline{\bigcup \overline{V_I}} = \overline{\bigcup V_I}$$
Hence, we always have
$$
\begin{aligned}
\partial \left(\bigcup V_I\right) & = \overline{\bigcup V_I} \setminus \bigcup V_I \\
{} & = \overline{\bigcup \overline{V_I}} \setminus \bigcup V_I \\
{} & \supset \overline{\bigcup \partial V_I} \setminus \bigcup V_I 
\end{aligned}
$$
Theorem A
Let $X$ be a topological space, $U, V \subset X$ both be open, and $U$ be connected. Then $U \cap V \neq \emptyset$ and $U \setminus V \neq \emptyset$ if and only if $U \cap \partial V \neq \emptyset$.
Proof A
Suppose $U \cap \partial V = \emptyset$. Then $U = (U \cap V) \cup (U \setminus \overline{V})$, and these subsets are disjoint. Since $U$ is connected, either $U \cap V = \emptyset$, or $U \setminus \overline{V} = \emptyset$. Because of the assumption, the latter is equivalent to $U \setminus V = \emptyset$. Suppose $U \cap \partial V \neq \emptyset$. Then $U \cap \overline{V} \cap \overline{X \setminus V} \neq \emptyset$, which implies $U \cap \overline{V} \neq \emptyset$ and $U \cap \overline{X \setminus V} \neq \emptyset$. Since $U$ is open, $U \cap \overline{V} = \emptyset \iff U \cap V \neq \emptyset$. Since $V$ is open, $U \setminus V = U \cap (X \setminus V) = U \cap \overline{X \setminus V} \neq \emptyset$. 
Theorem B
Let $(X, \mathcal{T})$ be a locally connected topological space, and $V_I$ be as in the problem description. Then $(*)$ holds.
Proof B
Let $U = \bigcup V_I$, and denote by $\mathcal{T}^*(x)$ the connected open neighborhoods of $x$. By Theorem A,
$$
\begin{aligned}
{} & x \in \partial U \\
\iff & x \in \overline{U} \setminus U \\
\iff & x \in \overline{\bigcup V_I} \setminus U \\
\iff & \forall W \in \mathcal{T}^*(x) : W \cap \bigcup V_I \neq \emptyset \land x \in X \setminus U \\
\iff & \forall W \in \mathcal{T}^*(x) : \exists i \in I: W \cap V_i \neq \emptyset \land x \in X \setminus U \\
\iff & \forall W \in \mathcal{T}^*(x) : \exists i \in I: W \cap V_i \neq \emptyset \land W \setminus V_i \neq \emptyset \land x \in X \setminus U \\
\iff & \forall W \in \mathcal{T}^*(x) : \exists i \in I: W \cap \partial V_i \neq \emptyset \land x \in X \setminus U \\
\iff & x \in \overline{\bigcup \partial V_I} \setminus U.
\end{aligned}
$$
Theorem C
Let $(X, \mathcal{T})$ be a topological space, and $V_I$ be as in the problem description, and also locally finite. Then $(*)$ holds.
Proof C
Let $U = \bigcup V_I$. For a locally finite collection (open subset or not), it holds that
$$\overline{\bigcup V_I} = \bigcup \overline{V_I}.$$ 
Therefore
$$
\begin{aligned}
\partial U & = \overline{U} \setminus U \\
{} & = \overline{\bigcup V_I} \setminus U \\
{} & = \bigcup \overline{V_I} \setminus U \\
{} & = \bigcup \{\overline{V_i} : i \in I\} \setminus U \\
{} & = \bigcup \{\overline{V_i} \setminus V_i : i \in I\} \setminus U \\
{} & = \bigcup \partial V_I \setminus U \\
{} & = \overline{\bigcup \partial V_I} \setminus U.
\end{aligned}
$$
 A: Here's a counterexample as per my comment. Let $X = 2^\omega$ be Cantor space with the usual topology, generated by basic clopen sets $[\sigma] = \{ \sigma^\frown \alpha: \alpha \in 2^\omega \}$ for finite strings $\sigma \in 2^{<\omega}$. Let $U \subseteq 2^\omega$ be any open set which is not closed (examples of such things here, e.g. the complement of a point). Then, $U = \bigcup_{i \in I} V_i$ for some basic $V_i$.
We use the fact that $A \subseteq X$ clopen $\iff$ $\partial A = \varnothing$. $U$ is not clopen, so $\partial U \neq \varnothing$. However, all the $V_i$ are clopen, so $\bigcup_{i \in I} \partial V_i = \varnothing$. It follows that $\partial U \neq \overline{\bigcup_{i \in I} \partial V_i} \setminus \bigcup_{i \in I} V_i$.
Presumably this works because Cantor space fails badly to be satisfy any sort of connectedness - it is totally disconnected.
A: Let $X$ be a zero-dimensional space (i.e. it has a base of clopen sets) and let $U$  be open in $X$ but not closed. 
Then write $U = \bigcup\{ U_i: i \in I\}$ where all $U_i$ are clopen. Then $\partial V_i = \emptyset$, so $\bigcup_i \partial V_i = \emptyset$ so the right hand side of your identity reduces to $\emptyset$ while te let hand side is just $\partial U \neq \emptyset$.
Such sets $U$ exist in $\Bbb Q$, $\Bbb R\setminus \Bbb Q$, the Cantor set, and in fact in most zero-dimensional spaces.
A: I'll show that the property (*) in the question implies locally connected, and hence that (*) is equivalent to local connectedness.
By contraposition, suppose $X$ is not locally connected. Then there exists $U \in \mathcal{T}$ such that $U$ is not sum-connected (meaning not all components are open). By this excellent recent answer, $\mathcal{T}|U \cap \mathcal{T}^C|U$ is not a topology, which in turn by the theorem in that question is equivalent to there existing $V \in_I \mathcal{T}|U$ such that $\partial_U {V_i} = \emptyset$, but $\partial_U (\bigcup V) \neq \emptyset$.
Then
\begin{eqnarray}
{} && (\overline{\bigcup \partial_X V_i}(X) \setminus \bigcup V) \cap U \\
= && \overline{\bigcup \partial_X V_i \cap U}(U) \setminus \bigcup V \\ 
= && \overline{\bigcup \partial_U V_i}(U) \setminus \bigcup V \\ 
= && \emptyset.
\end{eqnarray}
and
\begin{eqnarray}
{} && \partial_X (\bigcup V) \cap U \\
= && \partial_U (\bigcup V \cap U) \\
= && \partial_U (\bigcup V) \\
\neq && \emptyset.
\end{eqnarray}
Hence
$$\partial_X (\bigcup V) \neq \overline{\bigcup \partial_X V_i}(X) \setminus \bigcup V).$$
Above we used the facts that $\partial_X A \cap U = \partial_U (A \cap U)$ for any $A \subset X$ and $U \in \mathcal{T}$. Similarly $\overline{A}(X) \cap U = \overline{A \cap U}(U)$ for any $A \subset X$ and $U \in \mathcal{T}$.
