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Let $H$ be a Hilbert space and a closed operator $T$ defined on its domain $D(T)$ which is dense in H. Let $M$ be a closed subspace of $H$.

In Is intersection of a dense subspace and a closed subspace of a Hilbert space also Dense? it is answered "$D(T)\cap M$ can never be dense in $H$" (the statement isn't really precise, can we let$M=H$ and still holds?).

But is $D(T)\cap M$ dense in $M$?

Let $f\in M\subset H$, then there is $g\in D(T)$ such that for any $\epsilon>0$ we have $ \|f-g\|_H<\epsilon. $ But I am not sure if I can take $g$ from $D(T)\cap M$.

If not in general, when? (except for the trivial case $M\subset D(T)$) .

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    What if $v$ is in $H\setminus D(T)$ and $M=\mathbb C v$? I don't know about "when"? Edit: I clicked your link and see Martin's answer would also provide you a counterexample, a specific case of what I mentioned. – 2017-02-04
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    In light of the comment by @JonasMeyer (you beat me by $2$ seconds), you may want to add the hypothesis $D(T)\cap M$ is nontrivial. – 2017-02-04
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    @Aweygan: But then what if $v$ is as above and $0\neq w\in D(T)$, and $M=\mathrm{span}\{v,w\}$? – 2017-02-04
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    BTW, any subspace of a separable HIlbert space has a countable dense subset from general principles (this holds in all metric spaces, where separable is equivalent to having a countable open base). These examples just show that we cannot expect the same dense subset to work everywhere. – 2017-02-04

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This will fail as the fine example by Martin Agerami shows. There we have a dense subset $N$ of $H$ and a closed subspace $M$ such that $N \cap M = \{0\}$, so certainly not dense in $M$.

We can say it is dense though when we know that the interior of $M$ is non-empty, but in a normed space this implies that $M = H$ already. In topological terms, you cannot say more, $H \setminus D$, where $D$ is a dense subspace of $H$, might contain closed subspaces (like the span of finitely many vectors not in $D$) that $D$ is not dense in. We can only say $D$ restricted to open subsets is dense in that open subset, but as said, an open subspace is automatically the whole space, so that buys you very little.