Show that $(S^\perp)^\perp=\overline {\operatorname{span}(S)}$ . Let $H$ be a Hilbert Space. $S\subseteq H$ be a finite set .Show that $(S^\perp)^\perp=\overline {\operatorname{span} (S)}$ .
Now $\operatorname{span}(S)$ is the smallest set which contains $S$ and $\overline{\operatorname{span}(S)}$ is the smallest closed set containing  $S$. Also $S^{\perp\perp}$ is a closed set containing $S$. Thus $\overline{\operatorname{span (S)}}\subset S^{\perp\perp}$ .
How to do the reverse?

You don't need $S$ to be a finite set. This is true for any subset $S\subseteq \mathscr{H}$. 
First of all notice that $S^{\perp}$ is a closed subspace of $\mathscr{H}$ (using continuity of inner product) and $S \subseteq (S^{\perp})^{\perp}$ for any subset $S \subseteq \mathscr{H}$. Thus $Span(S) \subseteq (S^{\perp})^{\perp} \Rightarrow \overline{Span(S)} \subseteq (S^{\perp})^{\perp}$. Further notice that if $C \subseteq D$, then $D^{\perp} \subseteq C^{\perp}$. So, we have
\begin{align*}
 S \subseteq \overline{Span(S)} & & \Longrightarrow & & \left(\overline{Span(S)}\right)^{\perp} \subseteq S^{\perp} & & \Longrightarrow & & (S^{\perp})^{\perp} \subseteq \left(\left(\overline{Span(S)}\right)^{\perp}\right)^{\perp}.
\end{align*}
But $\left(\left(\overline{Span(S)}\right)^{\perp}\right)^{\perp} = \overline{Span(S)}$ (because if $M$ is a closed subspace of a Hilbert space $\mathscr{H}$, then $(M^{\perp})^{\perp} = M.$) Thus, $(S^{\perp})^{\perp} = \overline{Span(S)}.$
