We could have solved both (a) and (b) at once. Let $\mathfrak{T}$ be the family of all topologies on $X$ such that $d:X\times X\rightarrow\mathbb{R}$ is continuous. Then, $\mathfrak{T}$ is nonempty, as, for example, the discrete topology on $X$ is in $\mathfrak{T}$ . Further, for every subfamily of topologies $\{\mathcal{T}_{\alpha}\}\subset\mathfrak{T}$ , $\cap_{\alpha}\mathcal{T}_{\alpha}$ is a topology on $X$ (Exercise 4(a) of §13) such that $d$ is continuous (the preimage of every open set is open relative to every topology in $\{\mathcal{T}_{\alpha}\}$ , and, hence, relative to their intersection). Therefore, $\mathcal{T}=\cap_{\mathcal{U}\in\mathfrak{T}}\mathcal{U}$ is the coarsest topology relative to which $d$ is continuous. It remains to note, that $d$ is continuous relative to $\mathcal{U}$ iff $\mathcal{U}$ contains every ball $B_{d}(x,\epsilon)$ . Indeed, if $d$ is continuous, then it must be continuous in each variable, and, in particular, $B_{d}(x,\epsilon)=\{y|d(x,y)<\epsilon\}$ must be open for every $x$ . And if every ball $B_{d}(x,\epsilon)$ is open, then $(x,y)\in A=d^{-1}((a-\epsilon,a+\epsilon))$ implies $(x,y)\in B_{d}(x,\delta)\times B_{d}(y,\delta)\subset A$ where $\delta=\min\{\tfrac{a-d(x,y)+\epsilon}{2},\tfrac{d(x,y)-a+\epsilon}{2}\}$ , hence, $A$ is open, and $d$ is continuous. Therefore, $\mathcal{T}$ is the coarsest topology containing all balls $B_{d}(x,\epsilon)$ , which is the one generated by the balls as its basis (Exercise 5 of §13).