= Solution
{c}
If $x\in F$, using $y=x$ in the <infimum> gives $0\leq f(x)\leq0$, so $f(x)=0$.
Conversely, if $f(x)=0$, the definition of the <infimum> allows a choice $y_j\in F$ with $|x-y_j|<1/j$. The <sequence> $y_j$ is therefore a <convergent sequence> with <limit> $x$. Since $F$ is a <closed set>, its <limit> $x$ belongs to $F$. Hence
$$
\boxed{f(x)=0\iff x\in F.}
$$
Closedness matters: without it, zero <distance to a set> characterizes its <closure>, which can contain additional points.
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