I would appreciate some help verifying (probably explain since I am assuming that I have faulted) the following attempt trying to solve the following problem;
If $V$ is a dense subset of a metric space $W$ and suppose that every Cauchy sequence in $V$ converges to a point in $W$ - then, prove or disprove that $W$ is complete.
Proof. (Attempting that W is complete)
Let $(w_i)$ be a cauchy sequence in $W$. Further, express $(w_i)$ as
$v_1,w_1,v_2,w_2,...,v_n,w_n$
where all $v \in V$. For the "last" element mentioned, specifically $w_n$, it can either: $w_n \in V\subset W$ or $w_n \in W\setminus V$.
If $w_n \in V$, we can conclude that we have a Cauchy sequence within $V$ that converges to $w_n$.
If $w_n \in W\setminus V$ we conclude that we have a Cauchy sequence that converges to a point outside $V$ in the region of the complement of $V$ in $W$.
Since $V$ is a dense set, by definition we get that for all $\varepsilon >0$ and for all $w \in W$ there exists a $v \in V$ such that $D(v,w)<\varepsilon$.
Therefore, for a $w_n$ in this region we can find a $v_n$ in a converging Cauchy sequence (i.e $v_1,v_2,...v_n$) in $V$ such that $D(v_n,w_n)<\varepsilon$ which mean that
$v_1,v_2,...,v_n,w_n$ is a converging Cauchy sequence. Therefore, $W$ is complete. (QED?)
I am having a hard time reading your proposed proof. Sentences like:
"For the "last" element mentioned, specifically $w_{n}$, it can either: $w_{n}\in V\subseteq W$ or $w_{n}\in W\setminus V$."
Don't make much sense to me.
To show that $W$ is complete let $(w_{n})$ be a Cauchy sequence in $W$. Because $V$ is dense in $W$ there is, for each $n\in\mathbb{N}$, a $v_{n}\in V$ such that $d(w_{n},v_{n})<\frac{1}{n}$. You can then prove that $(v_{n})$ is a Cauchy sequence in $V$ (using the triangle inequality) that then must converge and that $(w_{n})$ must converge to the same limit.