Given a metric space $(X,d)$, how to prove that the function $d \colon X \times X \to \mathbf{R}$ is continuous?
If we take any two arbitrary real numbers $a$ and $b$ such that $a < b$, then we need to show that the set $d^{-1} (a,b)$ given by
$$ d^{-1} (a,b) := \{ (x, y) \in X \times X | a < d(x,y) < b \} $$
is open in the product topology on $X \times X$.
A basis for this product topology is the collection of all cartesian products of open balls in $(X,d)$.
For $a, b ∈ ℝ$, let $(x,y) ∈ d^{-1} (a..b)$, that is $a < d(x,y) < b$.
Now choose a real $ε > 0$ such that $U_{ε+ε} (d(x,y)) ⊆ (a..b)$ and look at $U_ε (x) × U_ε (y)$.
For any pair of points $(x',y') ∈ U_ε (x) × U_ε (y)$, one has \begin{align} d(x',y') &≤ d(x',x) + d(x,y) + d(y,y') \\ &< ε + d(x,y) + ε, \\ &\text{ – as well as –} \\ d(x,y) &≤ d(x,x') + d(x',y') + d(y',y) \\ &< ε + d(x',y') + ε, \end{align} which is to say that $d(x',y')$ and $d(x,y)$ are within $ε + ε$ from each other, or more formally $$\rvert d(x',y') - d(x,y) \rvert < ε + ε,$$ so $d(x',y') ∈ U_{ε+ε}(d(x,y)) ⊆ (a..b)$.
Thus, $U_{ε} (x) × U_{ε} (y) ⊆ d^{-1} (a..b)$.
Let $(x,y)\in d^{-1}(a,b)$. Define $\epsilon=\frac{1}{100}\min(d(x,y)-a,b-d(x,y))$. Then for any point $(\xi,\gamma)\in B(x,\epsilon)\times B(y,\epsilon)$, we have $${\color{blue} d(\xi,\gamma) }\leq d(\xi,x)+d(x,y)+d(y,\gamma)<2\epsilon+d(x,y)<{\color{blue}b}$$ and $${\color{blue} a}<d(x,y)-2\epsilon <d(\xi,\gamma)+\underbrace{ d(x,\xi)-\epsilon}_{<0} +\underbrace{d(y,\gamma)-\epsilon}_{<0}<{\color{blue} d(\xi,\gamma)}.$$ That is to say $$(x,y)\in B(x,\epsilon)\times B(y,\epsilon)\subseteq d^{-1}(a,b).$$
Hence any point $(x,y)\in d^{-1}(a,b)\subseteq X\times X$ is surrounded by some open set contained in $d^{-1}(a,b)$.
(After typing this answer, I realized this is EXACTLY the same as the answer by @k.stm; so thanks also to his contribution. I guess it will be hard to write a different one.)
