 Solution

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 $\epsilon >0$ with $A\cap B_{\epsilon }(b)\subseteq B$. 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, $B$ is closed in $A$, but not open in $A$. It is the intersection with the closed half-plane $\mathrm{Im}z\ge 0$. A relative neighborhood of zero contains points of negative imaginary part, so relative openness fails.