Distance Function of Metric Space is Continuous

Theorem
Let $\left( {M, d} \right)$ be a metric space.

Then for all $x, y \in M$, the mapping $\xi \mapsto d \left( {\xi, y} \right)$ is continuous in $x$.

Proof
Let $x_0, y \in M$.

Let $x \in M$ be such that $d \left( {x, x_0} \right) < \epsilon$.

Then by the reverse triangle inequality, $\left \vert { d \left( {x, y} \right) – d \left( {x_0, y} \right) } \right \vert \le d \left( {x, x_0} \right) < \epsilon$.

Hence the result by the $\epsilon$-$\delta$ definition of continuity (taking $\delta = \epsilon$).