I think that you can construct a counterexample if $X$ is the space of measurable functions on a nice probability space (like the unit interval with the Lebesgue measure) endowed with stochastic convergence where $y_n \to 0$ if and only if $P(|y_n| >\varepsilon) \to 0$ for all $\varepsilon >0$. Take a suitable sequence $A_n$ of measurable sets and (large) constants $c_n$ and define $r_n=c_n I_{A_n}$ (the characteristic function). Then $r_n \to 0$ whenever $P(A_n)\to 0$. Next define $y_n$ so that $r_n= 2y_{n+1}-y_n$ (namely $y_{n+1}=\sum\limits_{k=1}^n 2^{-(n-k+1)}r_k$ or something alike). Then you will see how to choose $A_n$ in order to avoid $y_n\to 0$.
Some more details: For $c_n= 2^{n+1}$ you get $$y_{2n+1} \ge \sum_{k=n}^{2n}c_k 2^{-(2n-k+1)}I_{A_n} \ge \sum_{k=n}^{2n} I_{A_k}.$$ Therefore $\bigcup_{k=n}^{2n} A_k$ is contained in $\lbrace |y_{2n+1}|\ge 1\rbrace$. If $A_n$ are independent with $P(A_n)=1/n$ you get $$ P(\lbrace |y_{2n+1}|\ge 1\rbrace) \ge P(\bigcup_{k=n}^{2n} A_k) = 1-\prod_{k=n}^{2n} (1-P(A_k)) = 1-\frac{n-1}{2n} \not\to 0$$