Let $X$ be a metric space, and let $x_0 \in X$ be a fixed point. Show that the real function $f_{x_0}$ given by $f_{x_0}(x) = d(x,x_0)$ is a continuous function.
The definition of continuity at a fixed point $x_0 \in X$ is $\forall \epsilon > 0, \exists \delta: d(x,x_0) < \delta \implies |f_{x_0}(x) - f_{x_0}(x_0)| < \epsilon$.
I know $f$ is continuous if it is continuous at every point in the domain ($X$), and I know that $f$ is continuous at a point if
Let $x \in X$ be an arbitrary point.
The RHS of the implication simplifies to $|f_{x_0}(x) - f_{x_0}(x_0)| = d(x,x_0)-d(x_0,x_0) = d(x,x_0) < \epsilon$.
So if $\delta = \epsilon$ the implication holds, and because we chose an arbitrary $x \in X$, $f$ is continuous everywhere .
Yes, this proof works for $x=x_0$.
But you only show continuity at $x_0$ but you need to do it for all $x$, so show the full non-contractiveness:
$$ \forall x, x' \in X: |f_{x_0}(x) - f_{x_0}(x')| \le d(x,x')$$
using the triangle inequality. Then $\delta=\varepsilon$ will still do.