Let $\varphi=(1+\sqrt{5}) / 2$ and $\psi=(1-\sqrt{5}) / 2$ be the roots of the quadratic equation $t^{2}-t-1=0$. So $\varphi \psi=-1, \varphi+\psi=1$ and $1+\psi=\psi^{2}$. An easy induction shows that the general term $c_{n}$ of the given sequence satisfies
$$c_{n}=\frac{\varphi^{n-1}-\psi^{n-1}}{\varphi-\psi} \quad \text { for } n \geq 0 \text {. }$$
Suppose that the numbers $\alpha$ and $\beta$ have the stated property, for appropriately chosen $m$ and $M$. Since $\left(c_{n}, c_{n-1}\right) \in S$ for each $n$, the expression
$$\alpha c_{n}+\beta c_{n-1}=\frac{\alpha}{\sqrt{5}}\left(\varphi^{n-1}-\psi^{n-1}\right)+\frac{\beta}{\sqrt{5}}\left(\varphi^{n-2}-\psi^{n-2}\right)=\frac{1}{\sqrt{5}}\left[(\alpha \varphi+\beta) \varphi^{n-2}-(\alpha \psi+\beta) \psi^{n-2}\right]$$
is bounded as $n$ grows to infinity. Because $\varphi>1$ and $-1<\psi<0$, the only way this can happen is if $\alpha \varphi+\beta=0$. Therefore, $\alpha \psi+\beta=\alpha(\psi-\varphi)=-\alpha \sqrt{5}$. The expression simplifies to
$$\alpha c_{n}+\beta c_{n-1}=\frac{-\alpha \sqrt{5}}{\sqrt{5}} \psi^{n-2}=-\alpha \psi^{n-2}.$$
Since $\psi^{n-2}$ is bounded, $\alpha c_{n}+\beta c_{n-1}$ is bounded if and only if $\alpha=0$. But this contradicts the condition of the lemma.

Let $j_{r}=j_{r+1}=1$; then $\sum_{r=1}^{k} \psi^{j_{r}}$ contains at least two summands equal to $\psi^{1}=\psi$. Like in the case $j_{r}=j_{r+1}=0$, we also infer that $j_{s} \neq 0$ and $j_{s} \neq 2$ for all $s$. Therefore
\psi x+y=\sum_{r=1}^{k} \psi^{j_{r}}0$. The current set $J$ should be \text { either } 1 \leq j_{1}\psi>-1$; therefore$x^{\prime}, y^{\prime} \geq 0$. Moreover, we have$3 x^{\prime}+2 y^{\prime}=2 x+y \leq \frac{2}{3} n$; therefore, if (3) holds then the induction applies: the numbers$x^{\prime}, y^{\prime}$are represented in the form as needed, hence$x, y$ also.

Now consider $\frac{\psi x+y-1}{\psi}$. Since
$$\frac{\psi x+y-1}{\psi}=x+(\psi-1)(y-1)=\psi(y-1)+(x-y+1)$$
we set $x^{\prime}=y-1$ and $y^{\prime}=x-y+1$. Again we require that $\frac{\psi x+y-1}{\psi} \in(-1, \varphi)$, i.e.
$$\psi x+y \in(\varphi \cdot \psi+1,(-1) \cdot \psi+1)=(0, \varphi) .$$
If (4) holds then $y-1 \geq \psi x+y-1>-1$ and $x-y+1 \geq-\psi x-y+1>-\varphi+1>-1$, therefore $x^{\prime}, y^{\prime} \geq 0$. Moreover, $3 x^{\prime}+2 y^{\prime}=2 x+y-1<\frac{2}{3} n$ and the induction works.

Finally, $(-1,-\psi) \cup(0, \varphi)=(-1, \varphi)$ so at least one of (3) and (4) holds and the induction step is justified.