= Solution
{c}
In the <subspace topology>, $B$ is relatively open in $A$ if $B=A\cap U$ for an open subset $U$ of the ambient Euclidean space. Equivalently, every $b\in B$ has a radius $\varepsilon>0$ with $A\cap B_\varepsilon(b)\subseteq B$. It is relatively closed if its complement $A\setminus B$ is relatively open, equivalently $B=A\cap F$ for an ambient closed set $F$.
For the upper half of the open disk, \b[$B$ is closed in $A$, but not open in $A$]. It is the intersection with the closed half-plane $\operatorname{Im}z\geq0$. A relative neighborhood of zero contains points of negative imaginary part, so relative openness fails.
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