Q:

Does there exist a ``continuous measure'' on a metric space?

Let $X$ be a separable complete metrizable space. Does there exist a complete metric $d$ and a Borel measure $\mu$ such that 
(a) 
$\mu(B_r(x))<\infty$ for every open ball $B_r(x)$ of radius $r>0$ around $x\in X$,
(b) for each $r>0$ the map
$
x\mapsto \mu(B_r(x))
$
is continuous on $X$,
(c) $\mathrm{supp}(\mu)=X$?

A:

By continuing the line of thought from Nate Eldredge's comment you can handle finitely many separated parts of the space by a distance having values at most one and exactly one between points in different connected components. However, a possible counter-example (I did not yet check the details) should follow by continuing this to countably many separated parts that accumulate. For instance:
Take $$X = [-1,0] \cup \{\frac1n\,:\,n \in \mathbb N\}$$ with the induced topology from $\mathbb R$. Each isolated point $\frac1n$ has to have positive measure. Since they accumulate to $0$ you cannot put a distance on $X$ separating them from the interval $[-1,0]$. This should force a discontinuity for a suitable $r>0$ when $x$ travels along $[-1,0]$.

