The answer is no. Let $U$ be any neighborhood of $0$ in $\mathbb{Q}$. Then, $\text{cl}_\mathbb{Q}\text{ }U=\mathbb{Q}\cap\text{cl}_\mathbb{R}\text{ }U$. Also, remember then that if $X,Y,Z$ are top. spaces and $X$ is a superspace of $Y$ then $Z$ will be a compact subspace of $Y$ if and only if it's a compact subspace of $X$ and thus for $\text{cl}_\mathbb{Q}\text{ }U$ to be compact in $\mathbb{Q}$ we need that $\mathbb{Q}\cap\text{cl}_\mathbb{R}\text{ }U$ is compact in $\mathbb{R}$. But, evidently this is not compact in $\mathbb{R}$, why?