Just a very quick argument which reduces the possibilities: Let $\Omega\subset{\mathbb R}\cup\{\pm\infty\}$ be the $\omega$-limit set of the sequence, that is the set of limits of "converging" sub-sequences. It is a non-void closed set by construction. The property $x_{a+1}-x_a\rightarrow0$ tells us that $\Omega$ is a connected set. The property $x_{2a}-x_a\rightarrow0$ tells us that $2\Omega=\Omega$. Therefore $\Omega$ can only be equal to one of the four sets $$\{0\},\quad[0,+\infty],\quad[-\infty,0],\quad{\mathbb R}.$$

Just a very quick argument which reduces the possibilities: Let $\Omega\subset{\mathbb R}\cup\{\pm\infty\}$ be the $\omega$-limit set of the sequence, that is the set of limits of "converging" sub-sequences. It is a non-void closed set by construction. The property $x_{a+1}-x_a\rightarrow0$ tells us that $\Omega$ is a connected set. The property $x_{2a}-2x_a\rightarrow0$ tells us that $2\Omega=\Omega$. Therefore $\Omega$ can only be equal to one of the four sets $$\{0\},\quad[0,+\infty],\quad[-\infty,0],\quad{\mathbb R}.$$

Edit. It was commented that the second property gives only an inclusion, of $2\Omega$ into $\Omega$. Actually, it does give also the reverse inclusion (hence the equality), when combined with the first property: Let $\ell$ be the limit of some subsequence $x_{n_k}$. Because of the first property, we may suppose that $n_k=2m_k$ is even. Then $\ell/2$ is the limit of $x_{m_k}$, hence $\ell/2\in\Omega$.