$\newcommand{\supp}{\mathop{supp}}$ The answer to the question as stated is negative, let $Y$ be a closed $\ell_{\infty}$ ball of radius $\varepsilon$ in $\mathbb{R}^d$ (equipped with metric $\ell_\infty$) and let $X$ be the interior of $Y$ (so that indeed $X$ is dense in $Y$).
Then $N(X, \varepsilon) = 1$ (we can cover $X$ by a single open ball in the origin). Whenever $X \subset Y$ we have $N(X, \varepsilon) \leq N(Y, \varepsilon)$, so this inequality is satisfied, but $N(Y, \varepsilon) \geq 2^d$ - all corners $\{\pm 1\}^d$ need to be covered by a different open ball.
On the other hand, when $X$ is dense in $Y$ we have for every $\varepsilon' > \varepsilon$ we have $N(Y, \varepsilon') \leq N(X, \varepsilon)$, which for many practical purposes is enough.
EDIT: As I understand now from comments, the constants $c$ and $C$ are not supposed to be universal -- they are allowed to depend on $X$ and $Y$. We will use the observation above, to show a counterexample for this statement as well, and moreover one where $Y$ is compact.
Indeed, let us consider $Y$ a disjoint union of $\frac{1}{k} B_\infty^{f(k)}$ over $k \in \mathbb{N}$ -- a $f(k)$-dimensional $\ell_\infty$ balls of radius $1/k$, and glue them along the common $0$. Now take $X$ --- a union of all interiors of $\frac{1}{k} B_{\infty}^{f(k)}$, for some fast growing function $f(k)$.
Concretely, taking $F(k) := \sum_{j < k} f(j)$, we can define $Y$ as a subset of $\ell_{\infty}$ given by $$ Y = \bigcup_k \{ x \in \ell_{\infty} : \supp(x) \subset [F(k), F(k) + f(k)), \|x\|_{\infty} \leq 1/k \}, $$ and $$ X = \bigcup_k \{ x \in \ell_{\infty} : \supp(x) \subset [F(k), F(k) + f(k)), \|x\|_{\infty} < 1/k \}, $$ with the induced metric from $\ell_{\infty}$. (Above $\supp(x) := \{ i : x_i \not= 0\}$.)
Then $$N(X, 1/k) = 1 + \sum_{j < k} N(\frac{1}{j} B_{\infty}^{f(j)}, 1/k) \leq 1 + \sum_{j < k} (2k)^{f(j)},$$ and $$N(Y, 1/k) \geq N(B_{\infty}^{f(k)}, 1) \geq 2^{f(k)}.$$
We can now pick $f(k)$ growing fast enough, so that for every possible $C$, there is $f(k)$ s.t. $$ 2^{f(k)} > C \sum_{j < k} (2k)^{f(j)}, $$ for example $f(1) = 1$ and for $k > 1$, $$ f(k) := \log_2 ( k \sum_{j < k} (2k)^{f(j)}). $$
