Let $M=\mathbb{R}$, with the standard metric, and let $N=(0,1] \subset M$. If $A=(0,1)$ then the closure of $A$ in $N$ is equal to $N$ (so $A$ is dense in $N$ ), but the closure of $A$ in $M$ is $[0,1]$.
Question: when we say the closure of $A$ in $N$ is equal to $N$, it seems that the closure of $A$ is not closed. But I also know the closure set is always closed. So I am confused.
I notice that your use of the closure terminology is careful: you say "the closure of $A$ in $N$", as you should.
But your use of the closed technology is not similarly careful. It should be!
You must not ever ask "Is $A$ closed?" in some kind of absolute sense. It's not like asking "Is $A$ finite?" which is an intrinsic property of $A$. Instead, closure is a property of a subset relative to a topological space or metric space like $M=\mathbb R$ or $N=[0,1)$. So, given $X$ and a subset $A$, you can ask "Is $A$ closed in $X$?" And if $A \subset B \subset X$ then you can ask "Is $A$ closed in $B$?".
So, for your problem, you can ask: "Is $A$ closed in $M=\mathbb R$?" or "Is $A$ closed in $N = [0,1)$?" or "Is $A$ closed in $A$?"
And the answers are: no, no, and yes.