Given $(X,p)$ is a metric space, we fix $a \in X$. I wish to show $f(x)=p(x,a)$ is uniformly continuous.
I think I have to work with the epsilon delta definition that is find a $\delta$ such that for every $\varepsilon$, $p(x,y)<\delta \implies f(x)-f(y)<\varepsilon$ but I am not sure how to do so..
Recall the reverse triangle inequality: $$|\rho(x, a), \rho(y, a)| \le \rho(x, y).$$ Then, if we fix $\varepsilon > 0$, we have, $$\rho(x, y) < \varepsilon \implies |f(x) - f(y)| = |\rho(x, a) - \rho(y, a)| < \varepsilon.$$ Therefore, $f$ is uniformly continuous (in fact, Lipschitz continuous).