If we take $X=\mathbb{R}$ and $Y=Z$ and $E=\mathbb{N}$ i.e. $\{1,2,3,4,5,\cdots\}$ then since for this case $E$ is open in $Y$ (as $Y$ is itself an entire metric space) however there does not exist any open set $G$ in $X$ for this particular set. then how $E$ is open in $Y$.
Pl clarify
If you want to be fancy, let $$U=\bigcup_{n\in\Bbb Z^+}\left(n-\frac12,n+\frac12\right)\;;$$ then $U$ is an open subset of $\Bbb R$, and $U\cap\Bbb Z=\Bbb Z^+$. But there’s no need to go to so much trouble, as Sean’s answer shows. For that matter, you could observe that $\Bbb Z_{\le0}=\Bbb Z\setminus\Bbb Z^+$ is a closed set in $\Bbb R$, so $\Bbb R\setminus\Bbb Z_{\le0}$ is an open set in $\Bbb R$ whose intersection with $\Bbb Z$ is clearly $\Bbb Z^+$.
