Open Sets in Metric Topology

Question

Munkres asks the following question:

Let $X$ be a metric space with $d$ as a metric. Show that $d: X \times X \rightarrow \mathbb{R}$ is continuous.

The solution here shows this by taking the preimage of an open set in $U \subset \mathbb{R}$, call it $d^{-1}(U)$, and constructing an open set in the domain that will map back into $d^{-1}(U)$. The conclude the proof here. How can one conclude $d^{-1}(U)$ is open by showing it contains an open ball? Many types of set have open sets in them - for instance a compact interval on the line.

Answer 1

What that proof shows is that for each $(x,y)\in d^{-1}(U)$, the set $d^{-1}(U)$ contains a product of open balls centered at $(x,y)$. By the definition of open set on a metric space, this proves that $d^{-1}(U)$ is open.

Answer 2

In fact for every $(x,y) \in d^{-1}[U]$ there are open sets $V_1,V_2$ of $X$ (of the form open balls), such that $(x,y) \in V_1 \times V_2 \subseteq d^{-1}[U]$. This makes $d^{-1}[U]$ an open set in the product topology (as all points are interior points of $d^{-1}[U]$).