Jan 30 at 2:00, by
0celo7 @ACuriousMind Define the set $K:=\{d(x_1,y)-d(x_2,y)\mid y\in A\}$. Then $D(x_1)-D(x_2)=\inf K$ (due to $\inf(A+B)=\inf A+\inf B$). It can be easily seen that for all $k\in K$, $k\le d(x_1,x_2)$. Then, by the previous lemma, $\inf K\le d(x_1,x_2)$. Due to symmetry, $|\inf K|\le d(x_1,x_2)$. Let $\epsilon >0$ be arbitrary and let $\delta =\epsilon$. Then $d(x_1,x_2)<\delta$ implies $|D(x_1)-D(x_2)|\le d(x_1,x_2) <\delta =\epsilon$, showing continuity of $D(x)$.