subset of binary space countable?

If you insist to have a bijection, here is one. Your set $S$ is the set of all binary infinite words containing no factor $11$. Any such word can be uniquely written as an infinite word on the alphabet $\{0, 10\}$, that is, $S = \{0, 10\}^\omega$. For instance, $$0100100010010(100)^\omega = \color{blue}{0}\color{red}{10}\color{blue}{0}\color{red}{10}\color{blue}{0}\color{blue}{0}\color{red}{10}0\color{red}{10}(\color{red}{10}\color{blue}{0})^\omega$$

It follows that the map from $\{0,1\}^\omega \to S$ consisting to replace every $1$ by $\color{red}{10}$ is a bijection. No need of the Cantor-Bernstein-Schroeder theorem.

You can explicitly write an injective function from the set of all binary sequences to the set you are considering: just map every sequence to one with $0$ inserted in every other position.

So no, this is definitely not countable.

Your set has an uncountable subset: consider the infinite sequences of $a$ and $b$ where $a=10$ and $b=00$, the set of these sequences has the same cardinality of the set of sequences of $0$ and $1$ (just map $a \mapsto 0$ and $b \mapsto 1$), and it is is clearly a subset of your set.