Let the function $d \colon \mathbb{R}^2 \longrightarrow \mathbb{R}$ be defined by$$d\big( (x, y) \big) := \lvert x-y \rvert \qquad \mbox{ for all } (x, y) \in \mathbb{R}^2. $$Let $\mathbb{R}$ and $\mathbb{R}^2$ have their usual topologies. Then how to show that this function $d$ is continuity by showing that the inverse image of every basic open set in the co-domain space $\mathbb{R}$ is an open set in the domain space $\mathbb{R}^2$?
My Attempt:
Let $u$ and $v$ be any real numbers such that $u < v$, and let us put$$] u, v[ := \{ r \in \mathbb{R} \colon u < r < v \}. $$Then we have$$\begin{align}d^{-1} \big( ]u, v[ \big) &= \left\{ (x, y) \in \mathbb{R}^2 \colon d \big( (x, y) \big) \in ]u, v[ \right\} \\&= \left\{ (x, y) \in \mathbb{R}^2 \colon u < d\big( (x, y) \big) < v \right\} \\&= \left\{ (x, y) \in \mathbb{R}^2 \colon u < \lvert x-y \rvert < v \right\} \\ &= \begin{cases} \left\{ (x, y) \in \mathbb{R}^2 \colon 0 \leq \lvert x-y \rvert < v \right\} \ \mbox{ if } u < 0 \\ \left\{ (x, y) \in \mathbb{R}^2 \colon u < \lvert x-y \rvert < v \right\} \ \mbox{otherwise} \end{cases} .\end{align}$$
How to proceed from here?
Of course, there are other, perhaps easier, proofs of this result using the methods of advanced calculus (i.e. elementary real analysis), but I would like to do it using these elementary and purely topological methods.