Solution: Let $(X,d)$ be a complete metric space and $Y \subseteq X$ be any subspace of $X$. We need to prove that $Y$ is complete if and only if it is closed. Let $Y$ be closed. Let $\left( y_n \right) $ be a Cauchy sequence in $Y$. We need to show that $(y_n)$ is convergent. Since $\left( y_n \right) $ is Cauchy in $Y$, it must be Cauchy in $X$, and $X$ is complete, so $y_n \rightarrow x \in X$. Since $Y$ is a closed set, and $y_n \to x$ implies $x \in Y$. Thus, the Cauchy sequence $\left( y_n \right) $ in $Y$ converges to $x$ in $Y$ and hence $Y$ is complete.

Now we assume that $Y$ is a complete metric space. We need to show that $Y$ is closed in $X$. Let $\left( y_n \right) $ be a sequence in $Y$ such that $y_n \to y$. We claim that $y \in Y$ which will prove that $Y$ is closed. Since $\left( y_n \right) $ is a convergent sequence, for any $\epsilon > 0$ we can find $n_0 \in \mathbb{N} $ such that for $n\geq n_0$ \[ d\left( y_n, y \right) \lt \frac{\epsilon}{2}. \] Therefore, for any $m \geq n_0$ we have $ d\left( y_n, y \right) \lt \frac{\epsilon}{2}$. Hence, for any $m,n \geq n_0$, we have \begin{align*} d\left( y_n, y_m \right) & \leq d\left( y_n , y \right) + d(y, y_m) \\ & \lt \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{align*} Hence $\left( y_n \right) $ is a Cauchy sequence in $Y$. As $Y$ is complete, $y_n \to y$ and the limit of a converging sequence is unique implies $y \in Y$. This completes the proof that $Y$ is closed.