Note that $(A^\perp)^\perp=\overline{\operatorname{Span}(A)}$ is equivalent to $(A^\perp)^\perp$ being the smallest closed subspace that contains $A$.

It is obvious that $ A\subseteq (A^\perp)^\perp $ and $ (A^\perp)^\perp $ is a closed subspace of the Hilbert space $ \mathcal{H} $. It remains to prove that $ (A^\perp)^\perp $ is the smallest one. Suppose $ F $ is another closed subspace containing $ A $, then $ A^\perp\supset F^\perp $ and hence $ (A^\perp)^\perp\subset (F^\perp)^\perp $. Now it suffices to show that for a closed subspace $ F $, $ (F^\perp)^\perp\subseteq F $(thus we have $ F=(F^\perp)^\perp $). And if you know this fact then we are done.

If you do not know this fact, here comes the proof:

Since a closed subspace $ F $ of the Hilbert space $ \mathcal{H} $ is also a Hilbert space and we know there exists an orthonormal basis for $ F $, say $ \{u_\alpha\}_{\alpha\in I} $ where $ I $ is an index set. We claim that there exists an orthogonal basis $ \{v_\beta\}_{\beta\in J} $, where $ J $ is an index set, for $ \mathcal{H} $ such that $ \{u_\alpha\}_{\alpha\in I}\subseteq\{v_\beta\}_{\beta\in J} $.

To prove this fact, consider the poset
$$ \mathcal{P}:=\{S\subsetneq\mathcal{H}: \{u_\alpha\}_{\alpha\in I}\subseteq S\text{ and }S\text{ is orthonormal}\}, $$

we know that $ \mathcal{P}\ne\emptyset $ and every chain has an upper bound in $ \mathcal{P} $ by taking their union, thus there exists a maximal element $ \tilde{S}\in\mathcal{P} $. If an element $ x\in\mathcal{H} $ satisfies $ \langle x,y\rangle=0 $ for any $ y\in\tilde{S} $, then by the maximality of $ \tilde{S} $, $ x $ has to be $ 0 $. Thus, by the definition of orthonormal basis of a Hilbert space, we know that $ \tilde{S} $ is an orthogonal basis for $ \mathcal{H} $. Write $ \tilde S=\{v_\beta\}_{\beta\in J} $ we know that $ \{v_\beta\}_{\beta\in J}\supseteq\{u_\alpha\}_{\alpha\in I} $.

Since every element $ x\in\mathcal{H} $ can be uniquely written as $$ x=\sum_{\alpha\in I}\langle x,u_\alpha\rangle u_\alpha+\sum_{\beta\in \{\beta\in J:v_\beta\notin\{u_\alpha\}_{\alpha\in I}\}}\langle x,v_\beta\rangle v_\beta, $$
it is clear that $ F^\perp=\operatorname{Span}(\{v_\beta\}_{\beta\in J}\setminus\{u_\alpha\}_{\alpha\in I}) $ and $ (F^\perp)^\perp=\operatorname{Span}(\{u_\alpha\}_{\alpha\in I})=F $ by the above expression and we are done.