Proving that $Y$ is complete Let $X$ be a dense subset of $Y$. 
Let every cauchy sequence in $X$ converge to a point in $Y$ $(1)$. 
By the definition of dense I know that every point in $Y$ is either in $X$ or is a limit point of $X$.
I aim to prove that $Y$ is compete, i.e that every cauchy sequence in $Y$ is convergent to a point in $Y$.
Consider an arbitrary cauchy sequence, either the sequence is completely in $X$, and then it follows from $(1)$, or the sequence is not completely in $X$ then there exists a $z\in Y-X$ in the sequence. Consider this sequence but without the element(s) in $Y-X$. This subsequence of the original sequence is completely in $X$ and thus converges to $Y$ by $(1)$. Now, I have a theorem stating that if a cauchy sequence has a subsequence convergent to $y$, then the original sequence converges to $y$. Thus the original cauchy sequence converges to a point in $Y$.
Is this correct? If it is not, how can I prove it?
 A: I'll elaborate a bit on my comment to show you why your approach doesn't quite work, and then give you a bit of a hint as to how to go forward.
Presumably you know that $\mathbb{Q}$ is dense in $\mathbb{R}$ and that every Cauchy sequence in $\mathbb{Q}$ converges to a point in $\mathbb{R}$. Consider the sequence $\{\frac{1}{\sqrt{2n^2}}\}$. This is a Cauchy sequence in $\mathbb{R}$ with no elements in $\mathbb{Q}$, and so your approach would result in an empty 'subsequence'.
Instead, note that if $X$ is dense in $Y$, for any $\epsilon>0$ and any $y\in Y$ there must be an $x\in X$ such that $\rho(x,y)<\epsilon$ (where $\rho$ is the metric on both $X$ and $Y$). If $\{y_n\}$ is your Cauchy sequence, consider the sequence that has a carefully chosen $x_n$ before each $y_n$ in the sequence. What can you say about this new sequence and what can you conclude from that?
A: If $(y_n)$ is a Cauchy sequence in $Y$ and for every $n$ we choose $x_n \in X$ with $d(x_n, y_n) < \frac{1}{2^n}$ by density of $X$ in $Y$, can you show that $(x_n)$ is also a Cauchy sequence in $X$? And that its limit $y \in Y$ (that exists by your assumption (1)) is also a limit for the original sequence $(y_n)$?
