I have as a definition
A lattice $L \subseteq \mathbb{R}^{n}$ is a subgroup that is free of rank $n$ such that $\mathbb{R}L = \mathbb{R}^{n}$.
I don't know if I am misinterpreting the statement, but taking $\mathbb{Z}^{2} \subseteq \mathbb{R}^{2}$, it doesn't seem to hold that $\mathbb{R}\mathbb{Z}^{2} = \mathbb{R}^{2}$. Wouldn't this imply that any point in $\mathbb{R}^{2}$ lies on a line through the origin with rational slope (so every ratio of real numbers gives a rational number)?
In you example, since $(0,1), (1,0) \in \Bbb{Z}^2$ you have $\Bbb{R}\Bbb{Z}^2 \supset \operatorname{span} \{ (0,1), (1,0) \} =\Bbb{R}^2$.
The symbol $\Bbb{R}L$ denotes the linear subspace spanned by $L$, so the condition $\Bbb{R}L = \Bbb{R}^n$ means that $L$ has $n$ $\Bbb{R}$-linearly independent vectors.