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 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 branches.) To see why this set is linearly indepedent, see this nice answer of Tim Gowers.