Prove that the set $\mathbb R\times\mathbb R$ is equipotent to the set $\mathbb R$? Please can proof the following statement?

$\mathbb R\times\mathbb R$ is equipotent to the set $\mathbb R$, where $\mathbb R$ is the real numbers.

 A: There clearly is an injection $\mathbb{R}\to\mathbb{R}\times\mathbb{R}$. By Cantor-Schröder-Bernstein, it suffices to find an injection $\mathbb{R}\times\mathbb{R}\to\mathbb{R}$, which is the same as finding an injection $(0,1)\times(0,1)\to\mathbb{R}$, because $\mathbb{R}$ is equipotent to $(0,1)$.
If $r\in(0,1)$, let $0.r_1r_2\dots r_n\dots$ be its unique decimal development (not eventually $9$, to have uniqueness, but eventually $0$ allowed).
To a pair $\langle r,s\rangle\in(0,1)\times(0,1)$, associate
$$
0.r_1s_1r_2s_2\dots r_ns_n\dots\in(0,1)
$$
The given decimal development cannot be eventually $9$ and from it it's possible to get back $r$ and $s$. So the mapping is injective.
A: Consider the following map:
Take the unique decimal representation of a Real (standard assumptions to get it unique, no infinite sequence of nines for instance.)
\begin{equation}
x=\sum_{k\in\mathbb{Z}}a_k\cdot 10^k
\end{equation}
and assign it the following vector in $\mathbb{R}^2$:
\begin{equation}
\left(\sum_{k\in\mathbb{Z}}a_{2k}\cdot 10^k,\sum_{k\in\mathbb{Z}}a_{2k-1}\cdot 10^k\right).
\end{equation}
EDIT the following is wrong (see Ujan's comment) but the mapping is still surjective which is enough to have:
"This clearly defines a bijection since you can easily write down the inverse."
