Completeness in metric spaces.

Let $$X$$ be a metric space and $$Y\subset X$$ a subspace of $$X.$$

I have to prove the statement:

(a) Prove that if $$X$$ is complete and $$Y$$ is closed, then $$Y$$ is complete.

(b) Prove that if $$Y$$ is complete then $$Y$$is closed in $$X$$.

I'm following the definition (Complete metric spaces). A metric space is called complete if every Cauchy sequence in the space converges.

But, I am confused how to start to prove....

For (a), if $$\{x_n\}$$ is a Cauchy sequence in $$Y$$, then it is a Cauchy sequence in $$X$$, thus converging to some point $$x\in X$$. Since $$Y$$ is closed, $$x\in Y$$ and so $$Y$$ is complete.
For (b), if $$\{x_n\}$$ is a sequence in $$Y$$ converging to some point $$x$$, we need to show that $$x\in Y$$. Since $$\{x_n\}$$ is a Cauchy sequence in $$Y$$, it converges to some point $$y\in Y$$. Then $$x=y$$ by the uniqueness of limits. Thus $$x\in Y$$ and so $$Y$$ is closed.
For (a), you can prove that $$Y$$ is complete using the definition of completeness. That is, take a sequence in $$Y$$, assume it is cauchy, and prove it converges.
1. $$Y$$ is closed if and only if every convergent sequence of elements from $$Y$$ has a limit in $$Y$$.
2. Every sequence of elements from $$Y$$ is also a sequence of elements from $$X$$.
For (b), you can prove that $$Y$$ is closed by proving that every convergent sequence in $$Y$$ has a limit in $$Y$$.