 Solution

The induced-metric balls are intersections with the corresponding ambient balls. On the horizontal arm the distance from the centre is $|x-1/2|$; on the vertical arm it is $\max (1/2,y)$. ConsequentlyThe closure relative to $E$ of the open ball is $[0,1]\times \{0\}$. Every point $(0,y)$ with $y>0$ has positive distance $y$ from that horizontal arm, so none of those points is in the closure. In particular $(0,1/4)$ lies in the closed ball but not in the closure of the open ball, proving strict inclusion.