(a) Let $(X,d)$ be a metric space, and let $(Y,d|_{Y\times Y})$ be a subspace of $(X,d)$. If $(Y, d|_{Y\times Y})$ is complete, then $Y$ must be closed in $X$.
(b) Suppose that $(X,d)$ is a complete metric space, and $Y$ is a closed subset of $X$. Then the subspace $(Y,d|_{Y\times Y})$ is also complete.
My solution
(a) Let us consider that $x\in X$ in an adherent point of $Y$. We have to prove that $x\in Y$.
Since $x$ is an adherent point of $Y$, there is a sequence $x_{n}\in Y$ which converges to $x\in X$.
Once $x_{n}$ is convergent, it is Cauchy and, due to the completeness of $Y$, it converges in $Y$.
Let us say that $x_{n}\to y\in Y$. Once limits are unique in metric spaces, we conclude that $x = y \in Y$.
Thus $Y$ contains all its adherent points and consequently it is closed.
(b) Let us consider a Cauchy sequence $x_{n}\in Y$. Since $Y\subseteq X$ and $X$ is complete, $x_{n}$ converges to $x\in X$.
Therefore $x$ is an adherent point of $Y$. Since $Y$ is closed, we conclude that $x_{n}\to x\in Y$.
Hence we conclude that $Y$ is complete.
Am I reasoning correctly? Is there another way to solve it?
Both reasonings are quite correct. I would have reasoned the same way.
Cauchyness does not depend on the surrounding space, just on the distances between the terms of the sequence; that's the basic realisation. And characterising closed sets by sequential closedness is a basic fact of metric spaces, as is unicity of limits.