Let's put ourselves in the framework of ZF. Is it true that if we think of the set of real numbers as a rational vector space, there are continuum-many linearly independent vectors? I feel that we could then use this to make an injection from $\mathbb{R}$ to $\mathbb{R}/\mathbb{Q}$, the quotient vector space by the 1-dimensional subspace $\mathbb{Q}$. Does this strategy work? This is to help me think about the result that (in ZF) $|\mathbb{R}| \leq |\mathbb{R}/\mathbb{Q}|$, proved by Mycielski in 1964. I haven't looked at his paper, just trying to imagine how one would prove this result. 



Continuum-many independent vectors over Q in R as a Q-vector space
Let $f: 2^{\mathbb{N}} \rightarrow \mathbb{R}/\mathbb{Q}$ be the function given by
$$ f((a\_i)\_{i \in \mathbb{N}}) = \text{the equivalence class of }\sum\_{k=0}^{\infty} \frac{b\_k}{2^{(k+1)!}}$$
where $(b\_i)\_{i \in \mathbb{N}}=(a\_0,a\_0,a\_1,a\_0,a\_1,a\_2,\dots)$. Notice that a real number of the form $$\sum\_{k=0}^{\infty} \frac{c\_k}{2^{(k+1)!}}$$ where $(c\_i)\_{i \in \mathbb{N}} \in 2^{\mathbb{N}}$ is rational if and only if the sequence $(c\_k)$ is eventually zero. Consequently, $f((a\_i)\_{i \in \mathbb{N}})=f((a'\_i)\_{i \in \mathbb{N}})$ if and only if the corresponding sequences $(b\_i)\_{i \in \mathbb{N}}$ and $(b'\_i)\_{i \in \mathbb{N}}$ are eventually equal if and only if $(a\_i)\_{i \in \mathbb{N}}=(a'\_i)\_{i \in \mathbb{N}}$. Finally, compose this function with your favorite explicit injection from $\mathbb{R}$ to $2^{\mathbb{N}}$. This gives you an injection from $\mathbb{R}$ to $\mathbb{R}/\mathbb{Q}$.


If you want to find a $\mathbb{Q}$-linearly independent subset of $\mathbb{R}$ of size continuum, consider the image of the map $g: \mathcal{A} \rightarrow \mathbb{R}$ given by
$$ g(S)=\displaystyle \sum\_{k=0}^{\infty} \frac{\chi\_{S}(k)}{2^{(k+1)!}}$$
where $\mathcal{A} \subseteq \mathcal{P}(\mathbb{N})$ is an almost disjoint family of size continuum. For example, enumerate the vertices of the full binary tree of height $\omega$ by $\mathbb{N}$ and let $\mathcal{A}$ be the set of (labels of) branches. To see why this set is linearly independent, see [this nice answer](https://mathoverflow.net/a/32780/33039) of Tim Gowers on another MO question.


Note that both of these constructions can be done in ZF.


