Let $(X,d)$ be a metric space and fix some $y \in X.$ Show that the function $f: X \to \Bbb R $ defined by $f(x) = d(x,y)$ is Lipschitz continuous. 
Let $(X,d)$ be a metric space and fix some $y \in X.$ Show that the function $f: X \to \Bbb R $ defined by $f(x) = d(x,y)$ is Lipschitz continuous.

Let $x, z \in X$. Using the triangle inequality one has $$f(x) = d(x,y) \leqslant d(x,z)+d(z,y ) =d(x,z) +f(z)$$
and $$f(z) = d(z,y) \leqslant d(z,x)+d(x,y) = d(x,z)+f(x).$$
Now I was trying to combining these to bound $|f(x)-f(z)|$ somehow, but this doesn't seem to work like I thought. I get that $|f(x)-f(z)| \leqslant |d(x,z)+f(z)-(d(x,z)+f(x))| =d(x,z) +f(z) +d(x,z)+f(x) = 2(d(x,z)) + f(x)+f(z)$, which doesn't help at all. What am I missing here?

$d(x,y) - d(z,y) \leq d(x,z)$ and $d(z,y) - d(x,y) \leq d(x,z)$
Thus
$|f(x) - f(z)| \leq d(z,x)$
