Injective function from $\mathbb{R}^2$ to $\mathbb{R}$?

Hint: Take the interval $[0, 1]$ and think how we might try to map pairs of numbers from that interval 1-1 into the interval by 'interleaving' binary representations. In other words if $a = \cdot a_1a_2a_3a_4\ldots$ and $b = \cdot b_1b_2b_3b_4\ldots$, then send $\langle a, b \rangle$ to $\cdot a_1b_1a_2b_2a_3b_3\dots$ ....

Let's construct an injective function $f : (0,1) \times(0,1) \to (0,1)$. Since there exist bijections between $\mathbb{R}$ and $(0, 1)$, the proposed function $f$ is sufficient to prove the existence of an injective function from $\mathbb{R}^2$ to $\mathbb{R}$.

Let the decimal representation of $x$ be $0.x_1x_2x_3\cdots$, and that of $y$ be $0.y_1y_2y_3\cdots$. Let $f(x, y)$ be $0.x_1y_1x_2y_2x_3y_3\cdots$.

To make this function well-defined, we should avoid decimal representations that end with infinite successive $9$s. Once this is taken care of, it's easy to show that this function is injective.

Whether or not this is sufficient will depend on what framework you are working in. The axiom of choice is equivalent to the statement that every surjective function $g$ has a right inverse. If you can find a surjection $g: \mathbb{R} \rightarrow \mathbb{R}^2$, you can just take $f$ to be a right inverse of $g$, which will necessarily be injective since $g \circ f$ is injective. If you don't have the axiom of choice, then you can't do this.

However, the assertion can be proved without appealing to the axiom of choice. As a hint, think about interweaving the digits of the number. You might have to do a bit of work to deal with non-uniqueness of decimal representations.