Here is an example in the spirit of the combinatorics of binary expansions. The moral is again that AC is not needed to exhibit uncountable linearly independent sets, even though it is needed to find bases.
Consider the family $\{ u_\alpha\} _ {\alpha\in\mathbb{R _ +}}, $ where $ u_\alpha $ is the real number whose binary sequence has support in the set $$S_\alpha:=\{ \lfloor \exp{\alpha k}\rfloor \ : \ k\in\mathbb{N} \}\ ,$$
namely $$u_\alpha:=\sum_{k \in S_\alpha} 2^{-k}\ .$$
This family is linearly independent over $\mathbb{Q}$. The relevant fact in order to see it is, that the subsets $S_\alpha\subset \mathbb{N}$ have the property that for any finite collection of them, say with $\alpha_1 < \alpha_2\dots < \alpha_r,$ the relative density of each of them, $S _ {\alpha_j},$ in their union $\cup_{1\leq i\leq r} S _ {\alpha_i}$ is exactly 1 if $j=1,$ and 0 otherwise (the smaller is $\alpha$, the thicker is $S_\alpha$). From this it follows easily that no non-trivial linear combination of $u_{\alpha_1},\dots,u_{\alpha_r}$ with integer coefficients may vanish (otherwise, one starts by looking at the coefficient relative to $u_{\alpha_1}$ and proves it has to be zero, otherwise $u_{\alpha_1}$ would be a linear combination of $u_{\alpha_2},\dots,u_{\alpha_r}$) u_{\alpha_2},\dots,u_{\alpha_r}$ with integer coefficients. This But this implies an inclusion of the supports, up to finitely many translations: $S_{\alpha _ 1} \subset \cup_{2\leq i\leq r} (S _ {\alpha_i}+F_i) $, for some finite sets $F_2,\dots,F_r$, contradicting the above stated density property).
