No, the proof is wrong.
Conceptual reason why it can't be correct: The argument in the paper, were it correct, would actually show much more, namely that for every bounded linear operator $T$ on a separable Hilbert space $H$ and every non-zero $x \in H$, the span of orbit $\{T^nx : n \in \mathbb{N}_0\}$ is not dense (since the choice of $x$ is arbitrary in the proof). This is obviously wrong (consider the right shift on $\ell^2$ and $x$ the first canonical unit vector).
What's wrong in the proof? The functional $f$ in formula (20) is not well-defined, in general, since the $y_n$ need not be linearly independent (and if they are, it's still not clear why it should be possible to extend $f$ continuously to all of $H$).
I don't know whether I have overlooked further errors in the proof.
Minor remark. There are a few further odd things about Theorem 1: The authors don't use that the underlying scalar field is complex. But over the real field the claim is wrong even in two dimensions. By the way, the statement of Theorem 1 is also wrong over the complex field if $H$ is one- or zero-dimensional.