It is apparently that this holds in Hilbert space, but I can not prove this for general inner product space or find a counterexample. (The only not complete inner product space known to me is $L^2$ restricted in continuous functions.)
Orthogonal complement of orthogonal complement
Here give a example, but it seems don't meet the condition closeness.
By the way, Is there any other less complicated spaces equipped with inner product but not complete?
A very simple innner product space which is not Hilbert is the spacee $\phi$ of sequences with only finitely many non-zero terms endowed with the scalar product of $\ell^2$. Let $L$ be the kernel of the continuous linear functional $x\mapsto \sum\limits_{n=1}^\infty x_n/n$ (the scalar product with $(1/n)_{n\in\mathbb N} \in\ell^2\setminus \phi$). Then $L^\perp =\lbrace 0\rbrace$: If $y\neq 0$ there is a last index such that $y_n\neq 0$. Defining $x_n=y_n$, $x_{n+1} = - (1+1/n) \,y_n$ and $x_k=0$ for all other $k$ gives $x\in L$ and $\langle x,y\rangle = y_n^2\neq 0$, i.e., $y\notin L^\perp$.
