Proof that the function $f(x)=d(x,y)$ is uniformly continuous? Consider a metric space $(M, {\rm d})$ 
and $y$  fixed in $M$.  
I want to prove that the function $f$ defined by $f(x)\colon={\rm d}(x,y)$ is uniformly continuous. 
So I know that if this function is uniformly continuous, then for all $\epsilon>0$, there exists a $\delta>0$ such that if the ${\rm d}(x_1,x_2)<\delta$, then this implies the ${\rm d}(f(x_1),f(x_2))<\epsilon$. 
So for all epsilon greater than zero, there exists a delta greater than zero such that ${\rm d}(x_1,x_2)<\delta \implies |{\rm d}(x_1,y)-{\rm d}(x_2,y)|<\epsilon$.
I can't figure out how to manipulate this equation to solve for delta!! 
I tried using the triangle inequality but it gets me nowhere. Please help!
 A: You have to show that for fixed $\;m\in X$= the metric space, the function $\;f(x):=d(m,x)\;$ is unif. continuous, so you have to show that
$$\forall\,\epsilon>0\;\exists\,\delta >0\;\;s.t.\;\;d(x,y)<\delta\implies |f(x)-f(y)|=|d(m,x)-d(m,y)|<\epsilon$$
But all you have to do then is use the triangle inequality and some symmetry:
$$d(m,x)\le d(m,y)+d(y,x)\implies |d(m,x)-d(m,y)|\le d(x,y)$$
and take now $\;\delta=\epsilon$ .
A: Let $\epsilon > 0$, choose $\delta = \epsilon$. Take $x_1,x_2 \in M$ such that ${\rm d}(x_1,x_2) < \delta$. Then: $$|f(x_1)-f(x_2)| = |{\rm d}(x_1,y)-{\rm d}(x_2,y)| \leq {\rm d}(x_1,x_2) < \delta = \epsilon.$$

Obs.: by the triangle inequality, we have that $${\rm d}(x_1,y)\leq {\rm d}(x_1,x_2)+{\rm d}(x_2, y)\implies {\rm d}(x_1,y)-{\rm d}(x_2,y)\leq {\rm d}(x_1,x_2).$$
Swapping $x_1$ and $x_2$ you get: $ {\rm d}(x_2,y)-{\rm d}(x_1,y)\leq {\rm d}(x_2,x_1) = {\rm d}(x_1,x_2).$ This way, the inequality $|{\rm d}(x_1,y)-{\rm d}(x_2,y)| \leq {\rm d}(x_1,x_2)$ follows from the definition of absolute value.
