I have a function $f:X \rightarrow Y$, where $(X,d)$ and $(Y,d')$ are metric spaces. I need to prove that $f$ is uniformly continuous on $X$ iff for all $A, B \subseteq X$ such that $d(A, B)=0$ implies that $d'(f(A), f(B))=0$
Necessity is clear. I can't show the sufficiency. I tried by contradiction. Suppose that $f$ is not uniformly continuous. Then, for some $\epsilon_0$ there are sequences $a_n$ and $b_n$ such that $d(a_n, b_n)\rightarrow0$ and $d'(f(a_n), f(b_n))>\epsilon_0$. Therefore, I defined the sets $A$ and $B$ as the set that contains the $a_n$'s and $b_n$'s respectively. So $d(A,B)=0$. By hypothesis $d'(f(A), f(B))=0$. I can't use this fact.
I proved the continuity of the function on $X$, but it's so hard prove the uniform. I can't show this part.