Let $X$ be set set of sequences $\ldots x_{-1}x_{0}x_{1}\ldots$ such that $x_{i}\in\{0,1\}$ and every $1$ is followed by an even number of $0$'s. An example of such a sequence in $X$ is $\ldots0011001000011100\ldots$. Now, let $W_{n}$ be the set of finite words $w$ of length $n$ that occur in $X$ such that $w$ starts either with $00$ or $1$. I want to prove that $$|W_{n+2}|\geq|W_{n+1}|+|W_{n}|$$ for all $n\in\mathbb{N}$. Clearly, $|W_{1}|=1$, $|W_{2}|=3$ and $|W_{3}|=6$, which proves the induction base step.
My attempt: I noted that $\{1w \ | \ w\in W_{n+1}\}$ and $\{00w \ | \ w\in W_{n}\}$ are disjunct. So $$|\{1w \ | \ w\in W_{n+1}\}\cup\{00w \ | \ w\in W_{n}\}|=|W_{n+1}|+|W_{n}|.$$ However, it is not true that $$\{1w \ | \ w\in W_{n+1}\}\cup\{00w \ | \ w\in W_{n}\}$$ is a subset of $W_{n+2}$ (If this was true, then we would be done). For example: $0001\in W_{4}$ but $10001\notin W_{5}$. Moreover, this doesn't use the induction hypothesis. So this is were I got stuck.
Any suggestions are greatly appreciated!