This is the follow up question to this question (*)
According to page 2 of this link 1 and this link 2, $\mathbb{R^Z}$ (which is referred as $\mathbb{R^\infty}$ in link 1) has elements of the following form:
$$(y_k)_{k\mathbb{\in Z}}=(\dots y_{-1},y_0,y_{1}\dots)$$
Or more concisely:
$$y: \mathbb{Z}\rightarrow \mathbb{R}$$ where there exists $(y_k)$ that have infinite nonzero components
The author in link 1 then begin a heuristic argument showing how one can naively attempt to construct a Hamel Basis for $\mathbb{R^Z}$, before concluding the construction does not end and then he mentioned it can be proved that the Hamel Basis of it is uncountable.
Further down in page 3 and 4, he uses Zorn's Lemma to show that all vector spaces have a maximal spanning set, hence a (Hamel) basis (assuming Axiom of Choice)
Now from (*), we found that $\mathbb{R^Z}$ is uncountable. Based on the author's heuristic argument, it make me suspect that the only set that can span $\mathbb{R^Z}$ need all nonzero elements in $\mathbb{R^Z}$. This sounds reasonable since such set is uncountable, which matches the requirement of the Hamel basis of $\mathbb{R^Z}$. Thus the question boils down to
What are the approaches that one should try to construct a proof of the statement in the title, or find counterexamples to it?