This question is essentially the same as double Orthogonal complement is equal to topological closure, but I have fewer "known" results. (Specifically, the linked Q/A proves that $A^{\perp\perp}=\overline{A}$ for a linear subspace $A$, assuming that $A^{\perp\perp}=A$ for closed $A$ is already known. As far as I'm concerned that's just another way to phrase the problem.)
Prove: If $V$ is an inner product space over $\Bbb C$ and $A$ is a linear subspace of $V$, then the topological closure $\overline{A}$ is equal to the double orthocomplement $A^{\perp\perp}$. Equivalently, if $A$ is a closed subspace, then $A^{\perp\perp}=A$.
To recap the definitions: $A^\perp=\{x\in V:\forall y\in A,\langle x,y\rangle=0\}$, while $\overline{A}$ is the topological closure induced by the norm $||x||=\sqrt{\langle x,x\rangle}$.
The converse to this, $A^\perp$ is closed, is a simple application of continuity of the inner product operation; from this we can deduce $\overline{A^\perp}=A^\perp$ and one half of the goal: $\overline{A}\subseteq A^{\perp\perp}$. I have also managed to prove $\overline{A}^\perp=A^\perp$ by another continuity argument.
If we additionally assume that $V$ is complete (i.e. a Hilbert space) and use the projection theorem (if $A$ is closed then $V=A+A^\perp$), we can write $V=\overline{A}+A^\perp$, so that given $x\in A^{\perp\perp}$ we can decompose $x=u+v$ for $u\in \overline{A},v\in A^\perp$; then $0=\langle x,v\rangle=\langle u,v\rangle+\langle v,v\rangle=|v|^2$, so $v=0$ and $x\in\overline{A}$ as desired.