= Solution
Intersect the ambient balls with $E$. All points on the vertical arm have distance at least $1/2$ from the centre, so the open ball is
$$
\boxed{(0,1)\times\{0\}}.
$$
The closed ball also contains the endpoints of that segment and the lower half of the vertical arm:
$$
\boxed{([0,1]\times\{0\})\cup(\{0\}\times[0,1/2])}.
$$
The closure of the open ball in $E$ is just $[0,1]\times\{0\}$. Both endpoints are limits of points in the open segment, whereas a vertical point of positive height has a small enough neighbourhood disjoint from it. Thus every point $(0,y)$ with $0<y\leq1/2$ witnesses strict containment.
\Image[/past-exam-of-the-options-ii-course-of-the-university-of-oxford/2008-metric-balls.png]
{title=Ambient and relative maximum-norm balls}
{description=The square on the left is the ambient closed ball. On the right, the open ball is the horizontal segment without its endpoints; its closure omits the additional vertical part of the relative closed ball.}
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