[proofplan] Each direction has a short proof using the interplay between convergence and the Cauchy property. For part (1), a convergent sequence in $N$ is Cauchy in $N$; completeness of $N$ gives a limit in $N$, and uniqueness of limits forces every limit point of $N$ to lie in $N$. For part (2), a Cauchy sequence in a closed subspace of a complete space is Cauchy in the ambient space, which provides a limit; closedness ensures the limit lies in the subspace. [/proofplan] [step:Prove that a complete subspace is closed] Let $(x_n)$ be a sequence in $N$ converging to some $x \in X$. Every convergent sequence is Cauchy, so $(x_n)$ is a Cauchy sequence in $N$. Since $N$ is complete, $(x_n)$ converges to some $y \in N$. By uniqueness of limits in the metric space $(X, d)$ (which follows from the positivity axiom: if $x_n \to x$ and $x_n \to y$, then $d(x, y) \leq d(x, x_n) + d(x_n, y) \to 0$, giving $x = y$), we have $x = y \in N$. Since every convergent sequence in $N$ has its limit in $N$, the subspace $N$ is closed. [/step] [step:Prove that a closed subspace of a complete space is complete] Let $(x_n)$ be a Cauchy sequence in $N$. Since $N \subseteq X$ with the induced metric, $(x_n)$ is also a Cauchy sequence in $X$. Since $X$ is complete, $(x_n)$ converges to some $x \in X$. Since $N$ is closed and $(x_n) \subset N$ with $x_n \to x$, the limit $x$ belongs to $N$. Therefore $(x_n)$ converges in $N$, and $N$ is complete. [guided] The argument chains three facts, each contributing one step of the logical chain from "Cauchy in $N$" to "convergent in $N$." **(i) A Cauchy sequence in a subspace is Cauchy in the ambient space.** Let $(x_n)$ be Cauchy in $N$. For any $\varepsilon > 0$, there exists $K$ such that $d(x_m, x_n) < \varepsilon$ for all $m, n \geq K$. This is the same inequality whether we view $(x_n)$ in $N$ or in $X$, since the metric on $N$ is the restriction of $d$. So $(x_n)$ is Cauchy in $X$. **(ii) Completeness of the ambient space provides a limit.** Since $X$ is complete, every Cauchy sequence in $X$ converges. Therefore there exists $x \in X$ with $x_n \to x$. **(iii) Closedness pulls the limit into the subspace.** The sequence $(x_n)$ lies entirely in $N$ and converges to $x \in X$. Since $N$ is closed, it contains all limits of convergent sequences from $N$: $x \in N$. Therefore $x_n \to x$ with $x \in N$, proving that $(x_n)$ converges in $N$. Since the Cauchy sequence was arbitrary, $N$ is complete. [/guided] [/step]