You have the right idea. Such a sequence is determined by:
- The length of the sequence, which takes values in $\mathbb{N} \cup \{ \infty \}$;
- The position (counting from $1$) at which the sequence switches from $0$ to $1$, which takes values in $\mathbb{N} \cup \{ \infty \}$, where a value of $0$ corresponds to a sequence which is constantly zero, and a value of $\infty$ (or a value greater than the length of a finite sequence) corresponds to a sequence which is constantly one.
So you can surject $(\mathbb{N} \cup \{ \infty \}) \times (\mathbb{N} \cup \{ \infty \})$ onto your set, by mapping $(\ell, p)$ to the sequence with length $\ell$, which has $0$ in all positions up to and including position $p$ and $1$ in all positions afterwards.
(P.S. This post uses the convention that $0 \in \mathbb{N}$.)