Let $(X,d)$ be some metric space, and let $x_0 \in X$ be some fixed point. Denote by $\mathcal{T}_0$ the topology with respect to which $x \mapsto d(x_0,x)$ is continuous. My question is,
If $x_1 \neq x_0$, will the map $x \mapsto d(x_1,x)$ be continuous wrt $(X,\mathcal{T}_0)$?
I've tried to prove that it is true but had no luck, so I suspect that it is false. However, I'm having trouble coming up with a counterexample. In term of trying to find a counterexample, I considered the function $f_0 : \Bbb{R} \to [0,\infty)$ and the topology $\mathcal{T}_0$ wrt it is continuous, and then tried showing that $f_1 : \Bbb{R} \to [0,\infty)$ via $f(x) = |1-x|$ is not continuous wrt $(\Bbb{R},\mathcal{T}_0)$ by considering various pullbacks of closed sets. Sadly, I didn't have much luck. I could use some help sorting all of this out.
Going back to the setup at the beginning of this post, ultimately I am trying to prove that $\mathcal{T}_0$ is finer than the metric topology. If I could show that $x \mapsto d(x_1,x)$ is in fact continuous wrt $(X,\mathcal{T}_0)$, then I would have an easy proof that $\mathcal{T}_0$ is finer than the metric topology.
Here is an answer if you read the question verbatim: consider $\mathbb R$ with the usual metric. One topology which makes $x \to d(x,0)$ continuous is the collection of sets of the form $\{x:|x| \in U\}$ where $U$ is open in $\mathbb R$. The function $x \to d(x,1)$ is not continuous in this topology because $\{x: |x-1| <1\}=(0,2)$ is not of the form $\{x:|x| \in U\}$ for any open set $U$. [ Sets of the the latter type are all symmetric about $0$].