$H$ is a Hilbert spaces and $M \subset H$ so $M^{\bot\bot}$ prove that it is the smallest closed subspace that covers M.
($M^\bot=\{x\in X|x\bot M \}$ and $M \subset M^{\bot\bot}$ and $x\in M \Rightarrow x \bot M^\bot \Rightarrow x\in M^{\bot\bot}$ )
One thing you can do is show that $(M^{\perp}) ^{\perp}=cl(M)$, and by definition the closure of the set is the smallest closed set containing it, one thing that is help full to use is that for a closed subspace $B$ we have $(B^\perp)^{\perp}=B$ and since $cl M^{\perp}=M^{\perp}$, basically because the inner product is continuous, we will have that $(M^{\perp})^{\perp}=(clM^{\perp})^{\perp}=clM$.