= Solution
{c}
Suppose $E$ is a <closed set> and $p\notin E$. Then $p$ is not a <limit point> of $E$, so some punctured <neighbourhood> of $p$ misses $E$. As $p\notin E$, the entire <neighbourhood> misses $E$. Thus $\mathbb R\setminus E$ is an <open set>.
Conversely, suppose every point $p\notin E$ has a <neighbourhood> disjoint from $E$. No such point can be a <limit point> of $E$, so every <limit point> lies in $E$. This proves the equivalence between containing all <limit points> and having an complement that is an <open set>.
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