Distance from a compact subset need not be attained in a metric space?

Question

Suppose we have a metric space $(X,d)$. Let $S$ be a compact subset of $X$. Provide me with an example of $X$, and $S$ (closed and bounded in $X$) such that $$\min \{d(p_0,p): p \in S\}$$ does not exist, where $p_0 \in X$.

Answer 1

The function $x\mapsto d(p_0,x)$ is continuous on $X$, hence attains its minimum on $S$ since $S$ is compact.

Answer 2

I'll assume $S$ not empty.

Since the function $f\colon S\to\mathbb{R}$, $f(p)=d(p_0,p)$ is continuous, when $S$ is compact its image is compact, hence closed and bounded; therefore the image of $f$ contains its minimum.

If $S$ is only assumed to be closed and bounded, but not compact, the statement is not generally true. Consider $X=\{0\}\cup (1,2]$, with the metric induced by $\mathbb{R}$. Then $S=(1,2]$ is closed and bounded, but $$ \inf_{p\in S} d(0,p)=1 $$ and there's no $p\in S$ such that $d(0,p)=1$.