Let $S\subseteq \mathbb{R}^n$.
The support function of set $S$ is defined as the following
$$ \sigma_S(x)=\sup_{y \in S} x^Ty $$ where $x \in \mathbb{R}^n$.
Show that $\sigma_S(x)=\sigma_{\bar{S}}(x) \,\,\,\,\forall \,\,\,x \in \mathbb{R}^n$.
If the set $S$ were closed then $S=\bar{S}$ so we are done. For the case that $S \neq\bar{S}$ we need to prove the statement. My approach is to say that the closure of a set is the set of all supporting hyperplains (which I do not know how to prove!), then show that $\sigma_S(x)$ takes its value for every $x$ when $y$ is at the boundary of $S$.
It is obvious that $\sigma_S(x) \leq \sigma_{ \overline {S}} (x)$. For the other inequality take $y \in \overline {S}$ and choose a sequence $\{y_n\} \subset S$ such that $y_n \to y$. Since $x^{T}y_n \leq \sigma_S(x)$ for all $n$ and $x^{T}y_n \to x^{T}y$ we get $x^{T}y \leq \sigma_S(x)$. Taking supremum over $y$ we get $\sigma_{ \overline {S}} (x) \leq \sigma_S(x)$.