A very silly question, yet I'm not 100% I got the good answer: for some $z \in (0,\infty)$, are the sets
$E_{1,z}:=\{(x_1, x_2) \in \bar{\mathbb{R}}_+^2\setminus \{0\}: x_2=+\infty, \, x_1 > z\}$
and
$E_{2,z}:=\{(x_1, x_2) \in \bar{\mathbb{R}}_+^2\setminus \{0\}: x_1=+\infty, \, x_2 > z\}$
relatively compact subsets of $\bar{\mathbb{R}}_+^2\setminus \{0\}$, where $\bar{\mathbb{R}}_+^2= [0, \infty]^2$? I would be tempted to answer: YES, since lines define closed subsets of the plan. Yet, the fact that $E_{1,z}$ and $E_{2,z}$ are placed on the boundary doesn't make me confortable with such answer. Any final word on that?
Addendum: on the other hand, would you say that sets of the type
$F_{1,z}:=\{(x_1, x_2) \in \bar{\mathbb{R}}_+^2\setminus \{0\}: x_2=0, \, x_1 > z\}$
and
$F_{2,z}:=\{(x_1, x_2) \in \bar{\mathbb{R}}_+^2\setminus \{0\}: x_1=0, \, x_2 > z\}$
are relatively compact? Their closure is $\bar{F}_{1,z}=[z, \infty]\times\{0\}$ and $\bar{F}_{2,z}=\{0\} \times [z, \infty]$, are those compact in $\bar{\mathbb{R}}_+^2$? I would say once more YES, in line with answer given below to the upper question.
E$_1$ = (z,oo]×{oo} is not closed, not compact. oo = $+\infty$
It is homeomorphic to (a,1] for some a in (0,1).
Defining E$_1$ = [z,oo]×{oo} will make it closed and compact.