Solution: Let $G = \bigcap \{G_n : n \lt \omega\}$ where each set $G_n$ is a dense open subset of $\mathbb{R}$. Define a Cantor scheme $(I_s : s \in 2^{\mathord{\lt}\omega})$ by recursion as follows. Let $I_\emptyset$ be a non-degenerate closed interval such that $I_\emptyset \subset G_0$, which is possible because $G_0$ is open and dense. Given $I_s$ where $s \in 2^n$, let $I_{s^\frown 0}$ and $I_{s^\frown 1}$ be disjoint non-degenerate closed subintervals of $I_s$ such that $I_{s^\frown 0},I_{s^\frown 1} \subset G_{n+1}$, which is again possible because $G_{n+1}$ is open and dense. We may choose these subintervals so that they have width $\le 2^{-n}$.
Finally, define a function $f: 2^\omega \to \mathbb{R}$ by letting $f(x)$ be the unique element of $\bigcap \{I_{x \restriction n} : n \le \omega\}$. (Recall that the intersection of a $\subset$-decreasing sequence of closed intervals whose length approaches zero is a singleton.) This function maps into $G$ because for all $x \in 2^n$ we have $I_{x \restriction n} \subset G_n$ for all $n \lt \omega$. Moreover, it is injective because for all distinct points $x,y \in 2^\omega$, letting $n$ be least such that $x(n) \ne y(n)$, the images $f(x)$ and $f(y)$ are contained in the disjoint intervals $I_{x \restriction (n+1)}$ and $I_{y \restriction (n+1)}$ respectively.