Disclaimer: Though I don't need it anymore this is interesting in its own!
Is it true that if the orthogonal complement is trivial then the subset was dense:
$$A^\bot=\{0\}\implies\overline{A}=X$$
Moreover does it hold that a dense subset remains dense in the completion:
$$\overline{A}^X=X\implies\overline{A}^{\hat{X}}=\hat{X}$$
Of course the converses are both trivial and I suppose both fail in general. Besides for complete a.k.a. Hilbert spaces the first is just a consequence of the orthogonal decomposition for closed subspaces.
Best Answer
The second is of course true. If $A$ is dense in $X$, then it is also dense in $\hat X$. Just approximate $x\in\hat X$ by $x_n\in X$, $||x-x_n||\le 2^{-n}$ and then approximate each $x_n$ by $y_n\in A$, $||y_n-x_n||\le 2^{-n}$.
The first is false: let $X=C_0^\infty(R)$, $\hat X=L^2(R)$, and $A=\{f\in X|\int_a^bf(x)dx=0\}$. Then $A^\perp=\{0\}$ in $X$, because $1_{[a,b]}\notin X$, but $A$ is clearly not dense.