$x_n \in V_{n+1} \setminus V_n$, and $x_n \to p$. Show that $p \in \partial V$ 
Suppose $\{V_n\}_{n=1}^\infty$ is a collection of open sets in $(X, d)$ such that $V_{n} \subset V_{n+1}$. Let $\{x_n\}$ be a sequence such that $x_n \in V_{n+1} \setminus V_n$ and suppose $\{x_n\}$ converges to $p \in X$. Show that $p \in \partial V$ where $V = \cup_{n=1}^\infty V_n$. 

It is not difficult to visualize that $x_n$ converges to the boundary of $V$, but I do not know how to prove it. I know that a convergent sequence in $\overline{V}$ should converge to $p \in \overline{V}$. 
 A: First of all note that all the elements in the sequence indeed belong to $\overline{V}$, hence the limit is also in $\overline{V}$. Now we will show that $p\notin V$. Suppose $p\in V$. Then by the definition of $V$ there is some $k\in\mathbb{N}$ such that $p\in V_k$. Since $V_k$ is open there is some $\epsilon>0$ such that the ball $B_{\epsilon}(p)$ is contained in $V_k$. Now, we know that $x_n\to p$ and hence:
$\exists (n_0\in\mathbb{N})\forall (n\geq n_0)\ [d(x_n, p)<\epsilon]$
And hence for all $n\geq n_0$ we have $x_n\in B_{\epsilon}(p)\subseteq V_k$, which contradicts the definition of the sequence. (because if $n>k$ then $x_n\notin V_k$)
So we showed that $p\in\overline{V}\setminus V$, hence it must be a point at the boundary of $V$. 
A: Because the $V$ is the union of open sets, it is open and hence $V = \text{int}(V)$.
Since all $x_n$ lie in $V$, their limit $x$ lies in $\overline{V}$, as you yourself said.
All that remains to show is that $x\not\in V$.
To that end, a simple lemma is useful.


Lemma: For each $m\in\Bbb N$ with $1\leqslant m \leqslant n$, we have that $x_n\not\in V_m$.

Proof:
Because of the chain of inclusions of the $V$s, if $x_n\in V_m$ for some $m\leqslant n$ we would have that $x_n\in V_{n}$.
This would contradict $x_n\in V_{n+1}\setminus V_n$. $\,\,\square$

Now, suppose $x\in V$.
If that were the case, then there would be some $k$ with $x\in V_k$.
Each neighborhood of $x$ must contain infinitely many of the $x_n$.
Because $V_k$ is open, there is some neighborhood of $x$ contained entirely in $V_k$.
However, by the lemma, for all $n\geqslant k$, we have $x_n\not\in V_{k}$.
This contradiction shows that our supposition that $x\in V$ must be false, as we set out to prove.

This proof shows that the claim is true in a more general setting (topological).
Metrics are not needed here.
A: I started writing this answer before Mark finished his, but maybe it will still be helpful. 
Let $U \subset X$ be an open set containing $p$. 
Since $x_n \to p$, there exists $N \in \mathbb{N}$ such that $n \geq N$ implies $x_n \in U$. In particular, this means $x_N \in U \cap V_{N + 1} \subset U \cap V$. Thus $U \cap V \neq \emptyset$. 
Suppose for contradiction that $p \in V$. This means that $p \in V_k$ for some $k \in \mathbb{N}$. Since $V_k$ is open and $x_n \to p$, there exists $M \in \mathbb{N}$ such that $n \geq M$ implies $x_n \in V_k$. Set $K = \max\{M, k\}$. Then $x_K \in V_k$ and $x_K \in X \setminus V_K \subset X \setminus V_k$, which is absurd. We conclude that $p \not\in V$. Thus $U \cap (X \setminus V) \neq \emptyset$. 
We have shown that any open set containing $p$ intersects with $V$ and $X \setminus V$. Therefore, by definition, $p \in \partial V$. 
