Let $X$ be a complete metric space and $Y$ be a subspace of $X$. Prove that $Y$ is complete iff it is closed.
My attempt: Firstly, suppose that $Y$ is closed. We want to show that $Y$ is complete.
Let $\{y_{n}\}$ be a Cauchy sequence in $Y$.Since, $Y$ is a subspace of $X$. So $\{y_{n}\}$ be a Cauchy in $X$ too.
Since, $X$ is a complete metric space. This implies that every Caucht sequence in $X$ converges in $X$. This implies the Cauchy sequence $\{y_{n}\}$ converges to $y$(say).
Since $Y$ is closed; so it contains all of its limit points. This implies $y \in Y$.
This shows that every Cauchy sequence $Y$ is convergent. Hence, $Y$ is complete.
Converse: Suppose that $Y$ is complete. We want to show $Y$ is closed.
Let $\{y_{n}\}$ be a Cauchy sequence in $Y$.
Since $Y$ is complete. This implies that every Cauchy sequence in $Y$ converges to a limit (say $y$) and $y \in Y$.
This implies $\{y_{n}\}$ converges to $y$ and $y \in Y$. This implies $Y$ contains all of its limit points.
Hence, $Y$ is closed.
Can anyone suggest me, is idea correct to proof this result?
Your proof that $Y$ is closed isn’t complete: you haven’t actually shown that if $x\in X$ is an arbitrary limit point of $Y$, then $x\in Y$. For this you have to start with an arbitrary limit point $x$ of $Y$. $X$ is a metric space, so there is a sequence $\langle y_n:n\in\Bbb N\rangle$ in $Y$ that converges to $x$. This sequence is of course Cauchy, so it must converge to a point of $Y$. But in a metric space — indeed, in any Hausdorff space — a sequence cannot converge to two different points, so $x$ must be in $Y$. Thus, $Y$ contains all of its limit points and is therefore closed.