= Solution
{c}
Put $F=\mathbb R^n\setminus\Omega$. This is a <closed set>. First assume $F$ is nonempty. The <Heine-Borel theorem> makes the nonempty bounded <closed set> $K$ a <compact set>. By the preceding parts, $x\mapsto d(x,F)$ is a <continuous function> and is strictly positive at every $x\in K$, because $K\subset\Omega$.
The <extreme value theorem> gives $x_*\in K$ attaining its minimum. Therefore
$$
\boxed{\operatorname{dist}(K,F)=\min_{x\in K}d(x,F)=d(x_*,F)>0.}
$$
The equality with the <distance between two sets> follows by taking the two <infima> successively. Crucially, <compactness> supplies a point attaining the minimum; positivity at each point alone would not make an arbitrary <infimum> positive. This is the principle that a <compact subset of an open set has positive distance from its complement>.
If $\Omega=\mathbb R^n$, its <complement> is empty. With the usual extended-real convention $\inf\varnothing=+\infty$, the stated inequality holds in that case too.
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