Let $(X,d)$ and $(Y,p)$ and $f:X\rightarrow Y$ uniformly continuous. Prove that for every $y_1,y_2\in f(x) $ with $y_1\neq y_2$ its true that $$dist( f^{-1}(y_1), f^{-1}(y_1))>0$$ where $dist(A,B)=inf\left \{ d(x,y): x\in A,y\in B\right \}$ .
Because $f$ is uniformly continuous $X$ is compact and Subjective so for every $y\in f(x)\exists \: x:f(x)=y\Rightarrow f^{-1}(y)=x$ is enough to show that $dist(x_1,x_2)>0$ and then is enough to show that $d(x_1,x_2)>0$
let's suppose $d(x_1,x_2)=0$ from the axioms of a metric space $\Leftrightarrow x_1=x_2$
$\Rightarrow f^{-1}(y_1)=f^{-1}(y_2)\Rightarrow y_1=y_2$ which is a contradiction from our hypothesis.
Suppose that $\operatorname{dist}(f^{-1}(\{y_1\}),f^{-1}(\{y_2\})) = 0$, and let $\varepsilon>0$. Then, because $f$ is uniformly continuous, there is a $\delta>0$ such that for any $x_1,x_2 \in X$ with $d(x_1,x_2)<\delta$ we have $\rho(f(x_1),f(x_2)<\varepsilon$.
Now, since $\operatorname{dist}(f^{-1}(\{y_1\}),f^{-1}(\{y_2\}) = 0$ we can find $z_1 \in f^{-1}(\{y_1\})$ and $z_2 \in f^{-1}(\{y_2\})$ such that $d(z_1,z_2) < \delta$. Thus, $\rho(y_1,y_2) = \rho(f(z_1),f(z_2)) < \varepsilon$.
Hence, we showed that for any $\varepsilon>0$ we have $\rho(y_1,y_2) < \varepsilon$, which implies that $\rho(y_1,y_2) = 0$.