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Let $\mathbb{R}^\infty$ be the vector space of real sequences $x=(x_1,x_2,x_3,\dots)$ and let $W$ be the subspace of the sequences $y$ such that $y_n=y_{n-1}+y_{n-2}$ for $n \ge 3$. Which is its dimension?

I think it is 2, since $y_3=y_1+y_2$ (that are two independent parameters), $y_4=y_2+y_3=y_1+2y_2$; $y_5=y_3+y_4=2y_1+3y_2$, $y_6=3y_1+5y_2$ and so on. Moreover, the base will be $$ Span \{(1,0,1,1,2,3,\dots), (0,1,1,2,3,5,\dots) \} $$ i.e. we obtain in both cases the Fibonacci sequence. Is the exercise well done?

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    It seems odd, not to say misleading, to use "$\mathbb R$" for anything other than the real numbers. But, yes, your subspace has dimension two over the reals for precisely the reason you give. – 2017-02-01

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Each element of $W$ is determined by two initial values $y_0,y_1$. In other words it's determined by the value $(y_0,y_1)\in\mathbb{R}^2$. So there is a correspondence between a basis of $\mathbb{R}^2$ and a basis of $W$. For the basis $\{(1,0),(0,1)\}$ of $\mathbb{R}^2$ the corresponding basis of $W$ is the one you already wrote.