I am trying to show that the following statement is false: if $M$ is a linear subspace of a hilbert space $X$ such that $M^⊥$ = {0}, then $M$ is dense $M^⊥=\{0\}$.
My counterexample: the infinite sequence space $l^2$ over the reals is a closed linear subspace of $l^2$ over complex numbers. The former has orthogonal complement $\{0\}$ but is not dense in the latter.
Does this work?
Your counter-example does not work because you are using complex scalars and real sequences do not form a linear subspace.
Actually there is no such example. If $M$ is not dense then Hahn Banach theorem gives a non-zero continuous linear functional which is $0$ on $M$. By Riesz Theorem this means ther exist $y \in H$ such that $ \langle x, y \rangle=0$ for all $x \in M$ but $y \neq 0$. But the $y \in M^{\perp} $ and $y \neq 0$.
