Let $X \subset Y \subset Z$ be three vector spaces. Suppose that $Y$ and $Z$ are complete metric spaces and that the injective map $Y \rightarrow Z$ is continuous (but not neccessarily open). Now suppose that $X$ is dense in $Z$.
Under what conditions is $X$ dense in $Y$?
One condition that I came up with is that $Z$ is locally compact and the inclusion $Y \rightarrow Z$ is proper. In that case the following argument works. Let $y \in Y$ be arbitrary. Then there exists a sequence $\{x_{n}\}_{n \in \mathbb{N}} \in X$ which converges to $y$ with respect to the topology on $Z$. Now let $U \subset Z$ be a pre-compact open neighbourhood of $y$. For $n$ larger than $N$, say, we know that $x_{n} \in U$. By the assumption that the inclusion of $Y$ in $Z$ is proper, we know that $U \cap Y$ is pre-compact. The sequence $\{ x_{n} \}_{n > N}$ lies in $U \cap Y$, and hence has a subsequence convergent with respect to the topology on $Y$, which must converge to $y$. Hence, for any $y \in Y$ there exists a sequence in $X$ which converges to $y$, hence $X$ is dense in $Y$.
I am particularly interested in the case that $Z$ is an infinite dimensional Hilbert space and that $Y$ is a Fréchet space.
(Also, I think that in general it's not true that $X$ is dense in $Y$, but I haven't been able to come up with a counter-example, I'd be happy to know one if it exists.)
A simple counterexample: $Y=\ell^1$ is dense a dense subspace of $Z=\ell^2$. Let $X=\{x\in\ell^1: \sum\limits_{n=1}^\infty x_n=0\}$ (the kernel of a linear functional on $Y$ without continuous extension to $Z$). Then $X$ is closed in $Y$ and still dense in $Z$:
Given $z\in\ell^2$ and $\varepsilon>0$ take $N$ large so that $\sum\limits_{n>N}|y_n|^2$ is very small. For $a=\sum\limits_{n=1}^N y_n$ the approximating element in $X$ can be taken as $x=(y_1,\ldots,y_N,-a/N^2,\ldots,-a/N^2,0,\ldots)$ with $N^2$ terms $-a/N^2$. (To see the approximation use Cauchy-Schwartz to estimate $|a|\le \|y\|_2 \sqrt{N}$.)