Exercise 4.3.12

Let $x\in R$ and let $a\in F$ be the element of $F$ closest to $x$ (must exist since $F$ is closed), we have $0\le g(y)\le |y-a|$ and $g(x)=|x-a|$. Similarly, let $b\in F$ be the element of $F$ closest to $y$. For the rest of the argument to make sense, we need to pick $a$ as our comparison point if $|x-a|\le |y-b|$ or $b$ otherwise. Suppose, without loss of generality, that we pick $a$. Thus, not only $g(y)-g(x)\le |y-a|-|x-a|$ but also $|x-a|\le |y-b|⟹0\le g(y)-g(x)$. This allows us to write

Applying the bound from Exercise 1.2.6 (d) we get

Setting $\delta =𝜖$ gives $|g(x)-g(y)|<𝜖$ as desired. If we had to pick $b$, the argument is the same just replacing $y$ with $x$ and vice-versa.

To see $g(x)\ne 0$ for $x\notin F$ notice that $F^c$ is open so there exists an $\alpha >0$ so that $V_{\alpha }(x)\cap F=\varnothing$ meaning $g(x)\ge \alpha$ and so $g(x)\ne 0$.