= Solution
{c}
The <distance to a set> is nonnegative and finite here: any fixed $y_0\in F$ gives $0\leq f(x)\leq|x-y_0|$. For every $y\in F$, the <triangle inequality> gives
$$
|x-y|\leq|x-x'|+|x'-y|.
$$
Taking the <infimum> over $y\in F$ yields $f(x)\leq|x-x'|+f(x')$. Interchanging $x$ and $x'$ proves
$$
\boxed{|f(x)-f(x')|\leq|x-x'|.}
$$
Thus $f$ is a <Lipschitz function> with <Lipschitz constant> one, so it is <continuous>. In the epsilon-delta definition of <continuity>, choose $\delta=\varepsilon$.
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