If $A$ is a compact subset of a metric space $(X,d)$ then A is closed and bounded.
What I'm confused about is this part of the proof:
Let $x_0$ be fixed and define the mapping $f:(A,T) \rightarrow \Bbb R$ by $f(a) = d(a,x_0)$ for all $a \in A$, where $T$ is the induced topology on A. Then $f$ is continuous.
How is f continuous?