 Solution

The distance to a set is nonnegative and finite here: any fixed $y_0\in F$ gives $0\le f(x)\le |x-y_0|$. For every $y\in F$, the triangle inequality givesTaking the infimum over $y\in F$ yields $f(x)\le |x-x^{\prime }|+f(x^{\prime })$. Interchanging $x$ and $x^{\prime }$ provesThus $f$ is a Lipschitz function with Lipschitz constant one, so it is continuous. In the epsilon-delta definition of continuity, choose $\delta =\epsilon$.