Assume that $N$ is a metric subsapce of $M$ and also is a closed subset of $M.$ A set $L \subset N$ is closed in $N$ if and only if it is closed in $M.$ Similarly, if $N$ is a metric subspace of $M$ and also is an open subset of $M$ then $U \subset N$ is open in $N$ if and only if it is open in $M.$
First, suppose $N$ is a closed metric subspace of $M.$ Consider a subset $L \subset N$ such that $L$ is closed in $M.$ Since $N$ is a closed in $M$ and $L = L \cap N,$ $L$ is closed in $N$ by the inheritance principle. Now, assume that $L$ is closed in $N.$ As $L \subset N$ = $L \subset M,$ it follows immediately that $L$ is closed in $M.$ Suppose that $N$ is an open metric subspace of $M.$ Consider a subset $U \subset N$ such that $U$ is open in $M.$ Since $N$ is open in $M$ and $U = U \cap N,$ $U$ is open by the inheritance principle. Now, assume that $U$ is open in $N.$ As $U \subset N$ = $U \subset M,$ it follows immediately that $U$ is open in $M.$
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I feel as if the question (given in a textbook) is almost pointless it is so trivial, and therefore wonder if my simple proof is insufficient as maybe I am missing something key. Can you verify it?
The second part of the closed section is too loose.
If A is closed within N, then exists M.closed K with A = K $\cap$ N.
Since both K and N are M.closed, A is M.closed.
Same problem persists in the second part of the open section.