If $X\subset\mathbb{R}^{d}$ lies in the closed half-space $\mathbb{h}^{+}$, then the closure $\text{cl} X$ of $X$ also lies in $\mathbb{h}^{+}$.
I understand the question but I don't have any idea how to prove this. Any hints/ help would be great. Thanks.
The closure of a set $X$ is (depending on your textbook: by definition) the smallest closed set containing $X$. In fact, such a smallest closed set is guaranteed to exist because the intersection of arbitrarily many closed sets is closed; and that means that the closure is just the intersection of all closed sets containing $X$. In particular, if $h^+$ is one of the closed sets containing $X$, then $h^+$ is one of the sets being intersected to form the closure, i.e., the closure is a subset of $h^+$.