Double orthogonal complement is equal to topological closure

So I'm in an advanced Linear Algebra class and we just moved into Hilbert spaces and such, and I'm struggling with this question.

Let $A$ be a nonempty subset of a Hilbert space $H$. Denote by $\operatorname{span}(A)$ the linear subspace of all finite linear combinations of vectors in $A$, and by $\overline{\operatorname{span}(A)}$ the topological closure of $\operatorname{span}(A)$ with respect to $\|\cdot\|$.

Also, let $A^⊥ = \{h ∈ H : \langle h,f \rangle = 0, ∀f ∈ A\}$ and $A^{⊥⊥} = (A^⊥)^⊥$. Use orthogonal projection to prove that $A^{⊥⊥} =\overline{\operatorname{span}(A)}$.

The thing that trips me up is that we don't know much about $A$, like if I knew a little more, perhaps to show it's closed, then I can do the direct sum decomposition blah blah, but it also confuses me why we're using the complement of $\operatorname{span}(A)$. Is it possible to show that $\operatorname{span}(A)$ is closed, then go from there? I know it might look a bit like a duplicate, but all the questions I find don't refer to orthogonal projection at all. Any hints would be greatly appreciated!

Let $U=\operatorname{span}(A)$. Then it's easy to see that $$ U^{\perp}=A^{\perp} $$ It's also easy to see that $$ \overline{U}^\perp=U^{\perp} $$ (where $\overline{U}$ denotes the closure of $U$) using continuity of the inner product. Thus $$ \overline{U}=\overline{U}^{\perp\perp}=U^{\perp\perp} $$ assuming you know that, for a closed subspace $V$, $V=V^{\perp\perp}$.