Let $x_1,x_2\in X$ and $r_1,r_2>0$ with $\|x_1-x_2\|\leq \varepsilon/2$ and $|r_1-r_2|\leq \varepsilon/2$. We will bound the distance between the two closed balls $C_1:=C_{x_1}(r_1)$ and $C_2:=C_{x_2}(r_2)$.
Let $w\in C_1$. If $w\in C_2$, then $\inf_{z\in B_2} \|w-z\|=0$. Let's assume that $w\notin C_2$, that is, $\|w-x_2\| > r_2$. Then we have $r_2\leq \|w-x_2\|\leq \|w-x_1\|+\|x_1-x_2\|\leq r_1+\varepsilon/2 \leq r_2+\varepsilon.$
Letting $w^*=x_2+{r_2\over \|w-x_2\|} (w-x_2)$ we have $w^*\in B_2$ and $\|w-w^*\|\leq \|w-x_2\|-r_2\leq \varepsilon.$ so $\inf_{z\in B_2} \|w-z\|\leq \varepsilon.$
Combining the two cases, and taking the supremum over $w\in C_1$, we deduce that $\sup_{w\in C_1}\inf_{z\in C_2} \|w-z\|\leq \varepsilon$. By symmetry we get $d_H(C_1,C_2)\leq \varepsilon$.
From this you can show that $(x,r)\mapsto B_x(r)$ is jointly continuous from $X\times (0,\infty)$ to $\mathcal{BC}(X)$. This is a bit stronger than the result you wanted.
Since $C_x(R)=x+C_0(R)$, your second question is answered here: How to prove that $f (x) = \mu (x + B)$ is measurable? assuming that $\mu$ is a non-negative, finite measure.