Let $(X,d)$ be a metric space and fix some $x_{0}\in X$. Show that the function $f(x) = d(x_{0},x)$ is Lipschitz continuous?
My problem wasn't with solving the question originally I did that using the triangle inequality and showing $\lvert f(x) - f(y)\rvert\le d(x,y)$, but the definition for Lipschitz continuity is $d_Y(f(x),f(y))\le K d_X(x,y)$ where $K > 0$, why can we assume that $d_Y$ is the the euclidean metric?
It is always the metric of your image space you need to look at. And since there was no other metric on $\mathbb{R}$ given, which is apparently the image space of $f$, one implicitly means the standard metric, i.e. the Euclidean one.