We will prove this theorem by demonstrating that for each point $x \ \times \ y \ \in \ X \ \times \ X$, and each open set $V$ around $d(x, y)$, there exists some neighbourhood $U$ of $x \ \times \ y$ such that $d(U) \ \in \ V$.

We begin by choosing an arbitrary point, $x \ \times \ y \ \in \ X \ \times \ X$. When we map this point under the metric, we get a real number, $d(x, \ y) \ \in \ \mathbb{R}$. Consider some open $V$ around this point. We know, from the definition of the standard/metric topology imposed on $\mathbb{R}$ that there exists some open ball around the point: $d(x, \ y) \ \in \ B(d(x, \ y), \ \epsilon)$. This is defined to be the set of all points $p$ of $\mathbb{R}$ such that: $$|p \ - \ d(x, y)| \ < \ \epsilon$$ It follows that $d^{-1}(B)$ is the set of all $a \ \times \ b$ such that: $$|d(a, b) \ - \ d(x, y)| \ < \ \epsilon$$ Now, let us switch gears and consider open sets around $x \ \times \ y$. Clearly, by definition of the product topology, the topology imposed on $X \ \times \ X$ will have a basis comprised of products of basis elements of $X$. Since the topology on $X$ is defined by the metric $d$, the basis elements will be open balls. Consider the following open balls around $x$ and $y$ in $X$: $$x \ \in \ B_{d}(x, \ \epsilon/2) \ \ \ \ \ \ \ \ \ \ \ \ y \ \in \ B_{d}(y, \ \epsilon/2)$$ Clearly, the set $x \ \times \ y \ \in \ B_{d}(x, \ \epsilon/2) \ \times \ B_{d}(y, \ \epsilon/2) \ = \ B_{d}$ is open in $X \ \times \ X$. Consider some element $a \ \times \ b \ \in \ B_{d}$. For this to be the case, it must be true that: $$d(a, \ x) \ < \ \epsilon/2 \ \ \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \ \ d(b, \ y) \ < \ \epsilon/2$$ By the triangle inequality, we have: $$d(x, \ y) \ - \ d(b, \ x) \ \leq \ d(b, \ y) \ < \ \epsilon/2$$ and $$d(b, \ x) \ - \ d(a, \ b) \ \leq \ d(a, \ x) \ \leq \ \epsilon/2$$ Combining these two equations, we get: $$d(x, \ y) \ - \ d(a, \ b) \ < \ \epsilon$$ This implies that $a \ \times \ b \ \in \ d^{-1}(B)$ for all $a \ \times \ b \ \in \ B_{d}$. It follows that $B_{d}$ is an open set of $X \ \times \ X$ such that $d(B_{d}) \ \subset \ B \ \subset \ V$ for all points $x \ \times \ y \ \in \ X \ \times \ X$. Thus, $d$ is continuous.