Bulgaria

Let $A=\{a_1,a_2,\dots ,a_n\}$ and set $$d(x,y)=\underset{1\le i\le k}{\max }|f_i(x)-f_i(y)|,g_i(x)=d(x,a_i)-d(a_i,a_i),1\le i\le n-1$$ and $d^{\prime }(x,y)={\max }_{1\le i\le n-1}|g_i(x)-g_i(y)|$. Since $$|g_i(x)-g_i(y)|=|d(x,a_i)-d(y,a_i)|\le d(x,y)(\text{why?})$$ then $d^{\prime }(x,y)\le d(x,y)$. On the other hand, $$d(a_j,a_i)=g_i(a_j)-g_i(a_i)\le d^{\prime }(a_j,a_i),1\le i\le n-1,1\le j\le n$$ $$ d(a_n, a_n) = 0 = d'(a_n, a_n) \text{ and hence } d'(x, y) = d(x, y).