 Solution

If $Y$ is closed and $(y_n)$ is Cauchy in $Y$, completeness of $X$ gives $y_n\to y\in X$, and closedness gives $y\in Y$. Conversely, if $Y$ is complete and $y$ lies in its closure, choose $y_n\in Y$ with $d(y_n,y)<1/n$. This sequence is Cauchy, so it converges in $Y$; uniqueness of limits makes its limit $y$. Thus $Y$ is complete exactly when it is closed in $X$.