I want to prove that given $f: (X,d) \rightarrow (X',d')$ uniformly continuous and $A,B \subseteq X: \ d(A,B) = 0$ then $d'(f(A),f(B))=0$.
Since $d(A,B) = 0$ we know that $\exists \ a \in A, b \in B: d(a,b) < \epsilon, \ \ \forall \epsilon > 0$. Since $f$ is uniformly continuous then $d'(f(a), f(b)) < \epsilon.$ Can I conclude from this that $d'(f(A), f(B)) = 0$?
I don't think I am using that $f$ is uniformly continuous here, but simply that it is continuous so I am probably missing something.
To show $d'(f(A), f(B))= 0$, we need to show that for all $\epsilon > 0$ there exist elements $x \in f(A)$ and $y \in f(B)$ such that $d'(x,y)<\epsilon$.
To this end, let $\epsilon >0$ be given. By uniform continuity of $f$, we may find a $\delta > 0$ such that whenever $d(a,b) < \delta$ we have that $d'(f(a),f(b)) < \epsilon$. Since $d(A,B)= 0$, we may find elements $a \in A$ and $b \in B$ such that $d(a,b) < \delta$. Then by the choice of $\delta$, taking $x = f(a)$ and $y = f(b)$ we obtain that $x \in f(A)$, $y \in f(B)$ and $d'(x,y) = d'(f(a),f(b)) < \epsilon$.
This doesn't work if $f$ is merely continuous but not uniformly continuous. For example, consider $X = X' = \mathbb{R}-\{0\}$ with the usual (Euclidean) metric (which we denote by $d$), and $f : X \to X'$ given by $f(x) = 1/x$. Then taking $A = (-1/2,0)$ and $B = (0,1/2)$ you can check that $d(A,B) =0$. However $f(A) = (-\infty, -2)$ and $f(B) = (2,\infty)$ so $d(f(A),f(B)) = 4$.