Fix a point $p=\langle x_0,y_0\rangle$, and suppose that $\epsilon\le 1$; what points are in $B(p,\epsilon)$, the open $\epsilon$-ball centred at $p$? A point $q=\langle x,y\rangle$ is in $B(p,\epsilon)$ iff $d(p,q)<\epsilon$. If $x=x_0$, $d(p,q)=|y-y_0|$, so we must have $y_0-\epsilon. If $x\ne x_0$, then $d(p,q)=1+|y-y_0|\ge 1\ge\epsilon$, so $q\notin B(p,\epsilon_0)$.
In other words, for $\epsilon\le 1$ the ball $B(p,\epsilon)$ is just the open vertical line segment
$\big\{\langle x_0,y\rangle:y_0-\epsilon
all points even moderately close to $p$ (meaning less than $1$ unit away) are on the same vertical line as $p$.
This means, for instance, that if $\langle q_n:n\in\Bbb N\rangle$ is a sequence converging to $p$, there must be an $m\in\Bbb N$ such that all of the points $q_n$ with $n\ge m$ lie on the line $x=x_0$: any point not on that line is at least $1$ unit away from $p$.
Now if $p=\langle x,0\rangle$, then $B(p,1)=\{\langle x,y\rangle:-1. Thus, $E=\bigcup_{x\in\Bbb R}B\big(\langle x,0\rangle,1\big)\;,$ and $\mathscr{B}=\left\{B\big(\langle x,0\rangle,1\big):x\in\Bbb R\right\}$ is an open cover of $E$. Does $\mathscr{B}$ have any proper subcover at all? Or is every single member of $\mathscr{B}$ necessary in order to cover $E$?