Is there a discrete rank-$\aleph_0$ free abelian subgroup of the topological group $\langle\mathbb{Z}_2 \times \mathbb{R},+\rangle$?
For example, denoting by $\mathbb{Z}$ the ring of integers and by $\mathbb{P}$ the set of prime numbers, the 'positive' square roots of the elements of $(1+8\mathbb{Z})\cap\mathbb{P}$:
$\bullet$ Exist in the $2$-adic integers $\mathbb{Z}_2$ by Hensel's lemma,
$\bullet$ Exist in the real numbers $\mathbb{R}$,
$\bullet$ Comprise a denumerable set by Dirichlet's theorem,
$\bullet$ Correspond bijectively between $\mathbb{Z}_2$ and $\mathbb{R}$ via $\{x^2 - p\in\mathbb{Z}[x]\colon p\in (1 + 8\mathbb{Z})\cap\mathbb{P}\}$,
$\bullet$ Are $\mathbb{Z}$-independent in the torsion-free additive group of the compact valuation domain $\mathbb{Z}_2$,
$\bullet$ Are $\mathbb{Z}$-independent in the torsion-free additive group of the locally compact field $\mathbb{R}$.
Enumerate these square roots $\{ b_k\cdot 1_{{\mathbb{Z}}_2}\}_{k\ge 1}$ and $\{ b_k\cdot 1_{\mathbb{R}}\}_{k\ge 1}$, respectively, where $1_{\mathbb{Z}_2}$ denotes the identity of $\mathbb{Z}_2$ and $1_\mathbb{R}$ denotes the identity of $\mathbb{R}$.
Is the rank-$\aleph_0$ free abelian subgroup $\big(\bigoplus\limits_{k\ge 1} \mathbb{Z}b_k\big)\cdot(1_{{\mathbb{Z}}_2},1_\mathbb{R})$ of $\langle\mathbb{Z}_2 \times \mathbb{R},+\rangle$ discrete?
There is not even a discrete free abelian subgroup of $\mathbb{Z}_2\times\mathbb{R}$ of rank $2$. Indeed, suppose $x,y\in\mathbb{Z}_2\times\mathbb{R}$ are $\mathbb{Z}$-linearly independent. For any $n$ and any $\epsilon>0$, there is a nontrivial linear combination of $2^nx$ and $2^ny$ whose $\mathbb{R}$-coordinate is within $\epsilon$ of $0$. Taking $n$ to be large, this linear combination will also have $\mathbb{Z}_2$-coordinate arbitrarily close to $0$. So, the subgroup generated by $x$ and $y$ contains nonzero elements in every neighborhood of $0$, so it is not discrete.